A=/X-12/ /Y 9/ 1997B= (x mũ 2 -16) /y-3/-2

Trần Khánh Linh

13 tháng 5 2018 lúc 10:17

bài 1: Rút gọn giá trị biểu thức:a) x(x+y) - y(x+y) với x(-1/2)mũ 5 : (1/2) mũ 4 và y8 mũ 2 : (-2) mũ 5b) (x-y) (x mũ 2 + xy + y mũ 2) -(x+y) ( x mũ 2 - y mũ 2 ) với x-y0c) x mũ 3 ( x mũ 2 - y mũ 2 ) + y mũ 2 ( x mũ 3 - y mũ 3 ) với x16 mũ 5 : 8 mũ 5 : (-2)mũ 4 và |y|1d) xy0; x y 1; x 1/2; y -3/2; x căn 4; y căn 9e) 5x ( 4x mũ 2 - 2x + 1) - 2x ( 10x mũ 2 - 5x-2) với x -3 ( -5 )g) 12- ( 2-3b ) + 35b - 9 ( b+1 ) với b (1/5) mũ 5 : (1/4) mũ 2f) ( x-y) ( x mũ 2 + xy + y mũ 2 ) + ( x+y ) ( x mũ 2...

Đọc tiếp

bài 1: Rút gọn giá trị biểu thức:

a) x(x+y) - y(x+y) với x=(-1/2)mũ 5 : (1/2) mũ 4 và y=8 mũ 2 : (-2) mũ 5

b) (x-y) (x mũ 2 + xy + y mũ 2) -(x+y) ( x mũ 2 - y mũ 2 ) với x-y=0

c) x mũ 3 ( x mũ 2 - y mũ 2 ) + y mũ 2 ( x mũ 3 - y mũ 3 ) với x=16 mũ 5 : 8 mũ 5 : (-2)mũ 4 và |y|=1

d) x=y=0; x = y = 1; x = 1/2; y= -3/2; x= căn 4; y= căn 9

e) 5x ( 4x mũ 2 - 2x + 1) - 2x ( 10x mũ 2 - 5x-2) với x = -3 ( -5 )

g) 12- ( 2-3b ) + 35b - 9 ( b+1 ) với b= (1/5) mũ 5 : (1/4) mũ 2

f) ( x-y) ( x mũ 2 + xy + y mũ 2 ) + ( x+y ) ( x mũ 2 -xy + y mũ 2 ) với x=2 và y = 2013 mũ 2014

Xem chi tiết

Lớp 8 Toán Câu hỏi của OLM

Trần Linh Nga

13 tháng 5 2018 lúc 9:25

bài 1: Rút gọn giá trị biểu thức: a) x(x+y) - y(x+y) với x(-1/2)mũ 5 : (1/2) mũ 4 và y8 mũ 2 : (-2) mũ 5 b) (x-y) (x mũ 2 + xy + y mũ 2) -(x+y) ( x mũ 2 - y mũ 2 ) với x-y0 c) x mũ 3 ( x mũ 2 - y mũ 2 ) + y mũ 2 ( x mũ 3 - y mũ 3 ) với x16 mũ 5 : 8 mũ 5 : (-2)mũ 4 và |y|1 d) xy0; x y 1; x 1/2; y -3/2; x căn 4; y căn 9 e) 5x ( 4x mũ 2 - 2x + 1) - 2x ( 10x mũ 2 - 5x-2) với x -3 ( -5 ) g) 12- ( 2-3b ) + 35b - 9 ( b+1 ) với b (1/5) mũ 5 : (1/4) mũ 2 f) ( x-y) ( x mũ 2 + xy + y mũ 2 ) + (...

Đọc tiếp

bài 1: Rút gọn giá trị biểu thức:

a) x(x+y) - y(x+y) với x=(-1/2)mũ 5 : (1/2) mũ 4 và y=8 mũ 2 : (-2) mũ 5

b) (x-y) (x mũ 2 + xy + y mũ 2) -(x+y) ( x mũ 2 - y mũ 2 ) với x-y=0

c) x mũ 3 ( x mũ 2 - y mũ 2 ) + y mũ 2 ( x mũ 3 - y mũ 3 ) với x=16 mũ 5 : 8 mũ 5 : (-2)mũ 4 và |y|=1

d) x=y=0; x = y = 1; x = 1/2; y= -3/2; x= căn 4; y= căn 9

e) 5x ( 4x mũ 2 - 2x + 1) - 2x ( 10x mũ 2 - 5x-2) với x = -3 ( -5 )

g) 12- ( 2-3b ) + 35b - 9 ( b+1 ) với b= (1/5) mũ 5 : (1/4) mũ 2

f) ( x-y) ( x mũ 2 + xy + y mũ 2 ) + ( x+y ) ( x mũ 2 -xy + y mũ 2 ) với x=2 và y = 2013 mũ 2014

Xem chi tiết

Lớp 8 Toán Bài 1: Nhân đơn thức với đa thức

ngonhuminh

13 tháng 5 2018 lúc 10:49

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4

Đúng 0

Bình luận (0)

quân phùng

17 tháng 5 2018 lúc 21:38

https://i.imgur.com/ZAuiaWv.jpg

Đúng 0

Bình luận (0)

Trịnh Thành Long K9A2 (...

9 tháng 7 2023 lúc 8:23

bài 4 : viết các biểu thức sau dưới dạng tích

a , 25 x mũ 2 -16

b , 16 a mũ 2 - 9 b mũ 2

c , 8 x mũ 3 + 1

d , 125 x mũ 3 + 27 y mũ 3

e , 8 x mũ 3 - 125

g , 27 x mũ 3 - y mũ 3

Xem chi tiết

Lớp 8 Toán Câu hỏi của OLM

Kiều Vũ Linh CTV

9 tháng 7 2023 lúc 8:32

a) 25x² - 16

= (5x)² - 4²

= (5x - 4)(5x + 4)

b) 16a² - 9b²

= (4a)² - (3b)²

= (4a - 3b)(4a + 3b)

c) 8x³ + 1

= (2x)³ + 1³

= (2x + 1)(4x² - 2x + 1)

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)(25x² - 15xy + 9y²)

e) 8x³ - 125

= (2x)³ - 5³

= (2x - 5)(4x² + 10x + 25)

g) 27x³ - y³

= (3x)³ - y³

= (3x - y)(9x² + 3xy + y²)

Đúng 2

Bình luận (0)

Nguyễn Đức Trí

9 tháng 7 2023 lúc 8:33

a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)

b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)

d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)

g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

Đúng 1

Bình luận (0)

Đoàn Khôi Nguyên

9 tháng 7 2023 lúc 8:39

tao phá olm đêy nè:

[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉φ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉ύ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉υ҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉β҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉η҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉[҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̅҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉̲҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉-҉]҉[҉̅҉̲҉҉҉]҉[҉̅҉̲҉]҉]҉[҉̅҉̲҉҉҉]҉

Đúng 0

Bình luận (0)

Lê Ngọc Thu Phương

25 tháng 7 2018 lúc 11:34

Làm phép chia:
a,(10 mũ 12 + 5 mũ 11 . 2 mũ 9 - 5 mũ 13 . 2 mũ 8) : 4 . 5 mũ 5 . 10 mũ 6
b,[5(x - y)mũ 4 - 3(x -y)mũ 3 + 4(x -y)mũ 2] : (y - x)mũ 2
c,[(x+y)mũ 5 - 2(x+y)mũ 4 + 3(x+y)mũ 3] : [-5(x+y)mũ 3]

Xem chi tiết

Lớp 8 Toán Bài 11: Chia đa thức cho đơn thức

Yukru

7 tháng 8 2018 lúc 20:33

a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)

\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)

\(=2^4.5+2-5^2\)

\(=57\)

b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)

\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)

c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)

\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)

\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)

Đúng 0

Bình luận (0)

Quỳnh Anh Phạm

16 tháng 12 2022 lúc 19:50

Tìm x , y biết

a ) x mũ 2 = x mũ 5

b ) ( 3x - 12 )mũ 15 = ( x - 17)mũ 15

c ) ( 4x - 16 )mũ 15 - ( x - 2 )mũ 15 = 0

d ) ( x - 3 )mũ 11 = ( 2x - 6 )mũ 11

Xem chi tiết

Lớp 6 Toán Câu hỏi của OLM

Đoàn Trần Quỳnh Hương

16 tháng 12 2022 lúc 19:56

bạn có thể check lại đề bài câu a được không ạ

Đúng 0

Bình luận (0)

Quỳnh Anh Phạm

16 tháng 12 2022 lúc 20:01

là x^2 = x^5 á bạn

Đúng 0

Bình luận (0)

Nguyễn Thị Hà Thảo

27 tháng 1 2023 lúc 14:49

x mũ 2/9 =y mũ 2 / 16 và x mũ 2 + y mũ 2 =100

Xem chi tiết

Lớp 7 Toán Câu hỏi của OLM

Dang Tung

27 tháng 1 2023 lúc 14:55

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\\ =>\left\{{}\begin{matrix}x^2=9.4=36\\y^2=4.16=64\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)

Đúng 3

Bình luận (0)

Đinh Thị Kim Huệ

30 tháng 7 2017 lúc 10:02

Tìm x y thuộc N:

a)(3x-12)*2y=0

b)(x mũ 2+3)*(y-3)=0

c)5 mũ 2x*2 mũ y+1=100

d)3 mũ x2 *5 mũ y-2=75

e)9 mũ x*49=144

Xem chi tiết

Lớp 6 Toán Câu hỏi của OLM

Thắng Nguyễn Đức

10 tháng 10 2021 lúc 8:17

bài 6 quy dồng mẫu thức các phân tử a)1 phần x+1 và 6 phần x-x mũ 2 với x khác 0 và x khác - hoặc + 1 b) y+5 phần y mũ 2 +8y +16 và y phần 3 y+12 với y khác -4

Xem chi tiết

Lớp 8 Toán Bài 9: Phân tích đa thức thành nhân tử bằng cách p...

Vũ Khánh Linh

5 tháng 9 2018 lúc 21:13

Tìm x , y , z biết

a)x/y=9/7 ; y,z=7/3 và x-y+z=-15

b)x/y=7/20 ; y/z=5/8 và 2x+5y-2z=100

c)x/12=y/9=z/5 và xyz=20

d)x/5=y/7=z/3 và x mũ 2 + y mũ 2 + z mũ 2=585

Xem chi tiết

Lớp 7 Toán Bài 7: Tỉ lệ thức

Chitanda Eru (Khối kiến...

18 tháng 9 2018 lúc 21:34

a) Từ \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

Từ \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\) (2)

Từ (1) và (2) =>\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3\cdot9\\y=-3\cdot7\\z=-3\cdot3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

b) Từ \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\) (1)

Từ \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\) (2)

Từ (1) và (2) =>\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=\dfrac{100}{50}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot7\\y=2\cdot20\\z=2\cdot32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=40\\z=64\end{matrix}\right.\)

c) Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)

=> \(x=12k\) ; \(y=9k\) ;\(z=5k\)

=> xyz = \(12k\cdot9k\cdot5k\) =\(540\cdot k^3\) = 20

=>\(k^3=20:540=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\)

=>\(k=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\cdot12\\y=\dfrac{1}{3}\cdot9\\z=\dfrac{1}{3}\cdot5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=\dfrac{5}{3}\end{matrix}\right.\)

d) Từ \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{25+49+9}=\dfrac{585}{83}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{585}{83}\cdot25\\y^2=\dfrac{585}{83}\cdot49\\z^2=\dfrac{585}{83}\cdot9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=\\y^2=\\z^2=\end{matrix}\right.\) đề bài sai nên ko tìm được x ; y ; z

Đúng 0

Bình luận (0)

Câu hỏi

Khoá học trên OLM (olm.vn)