a=x^3+16Y^2-5XY TAI X=16 Y=3
A= 5xy – 13x + 5y khi x = - 3 ; y = 2
B= – 2x + 16y – 15xy khi x = 2 ; y = - 1
A = 5 . ( - 3) .2 - 13 .( -3) + 5.2
A = - 30 + 39 + 10
A = 19
B = -2x + 16y - 15xy
B = (-2) . 2 + 16 .( - 1) - 15.2.(-1)
B = -4 - 16 + 30
B = 10
A = 5 . ( -3 ) . 2 - 13 . ( -3 ) + 5 . 2 B = -2 . 2 + 16 . ( -1 ) - 15 . 2 . ( -1 )
= ( -30 ) - ( -39 ) + 10 = -4 + ( -16 ) - ( -30 )
= -30 + 39 + 10 = ( -30 ) + 30
= 9 + 10 = 0
= 19
a,(x^2+x-1)^2-(x^2+2x+3)^2
b, -16+(x-3)^2
c, 64+16y+y^2
Viết biểu thức dưới dạng tích. Các bn giúp mình vs ạ
a: Ta có: \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
b: Ta có: \(\left(x-3\right)^2-16\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x+1\right)\left(x-7\right)\)
c: \(y^2+16y+64=\left(y+8\right)^2\)
Tinh gia tri cua bieu thuc
M=5xy-10=3y tai x=2 ; y=-3
N =2(x^2-1) +3x-2 tai x=-1
\(M=5.2.\left(-3\right)-10=3.\left(-3\right)\)
\(M=-30-10=-9\)
\(M=-40+9\)
\(M=-31\)
\(N=2\left(x^2-1\right)+3x-2\)
\(N=2.\left(1-1\right)+3.\left(-1\right)-2\)
\(N=-3-2\)
\(N=-5\)
Thay x= -1 vào biểu thức, ta được:
N= 2 .((-1)^2 -1) + 3.(-1) -2
= 2.(1-1) +(-3)-2
= 2.0 + (-3) -2
= 0 +(-3) -2
= -5
a) 5x^3-10x^2+5x-9y^2
c) x^3-3x^2+3x-y^3-1
d) (x+3)^2.y^2-16y^4
e) 8(x-1)^2y^2+8(1-x)^2y-16(x-1)^2
Dấu ni ^ là mũ nha
\(x^3-3x^2+3x-y^3-1\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right).\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\)
a) 16 - (x-3)^2
b) 64 + 16y + y^2
c) 1,24^2 - 0,24^2
d) 1/8 - 8x^3
e) 100 - (3x - y)^2
f) 64x^2 - (8x + 3)^2
Đề là gì bạn nhỉ?
\(16-\left(x-3\right)^2=4^2-\left(x-3\right)^2=\left(4-x-3\right)\left(4+x-3\right)\)
\(64+16y+y^2=y^2+2y4+4^2=\left(y+4\right)^2\)
\(1,24^2-0,24^2=\left(1,24-0,24\right)\left(1,24+0,24\right)=1.1,48=1,48\)
\(\frac{1}{8}-8x^3=\left(\frac{1}{2}\right)^3-\left(2x\right)^3=\left(\frac{1}{2}-2x\right)\left(\frac{1}{4}+x+4x^2\right)\)
\(100-\left(3x-y\right)^2=10^2-\left(3x-y\right)=\left(10-3x+y\right)\left(10+3x-y\right)\)
\(64x^2-\left(8x+3\right)^2\)
\(=\left(8x\right)^2-\left(8x+3\right)^2\)
\(=\left(8x-8x-3\right)\left(8x+8x+3\right)\)
\(=\left(-3\right)\left(16x+3\right)\)
\(=-48x-9\)
9 Phân tích đa thức sau thành nhân tử:
a) 9xy^2-18x^2y ; b) 6x^2-2y ; c)7x(x-y)-14y(y-x)
d)7-x^2 ; e) 16+8x+x^2 ; f)1-27x^3
g) x^3-9x^2+27x-27 ; h) (x+2y)^2-16y^2 ; i) x^3-64y^3
Tinh
Xy^3+4xy^3-3xy^3
(-4/5ab^2c)×(-20a^4b^3c)
Bai 2.tinh gia tri cua bieu thuc a=14x^2+5xy-2010y^2 tai x=-1;y=-2
xy3+4xy3-3xy3
=5xy3-3xy3 = 2xy3
tươg tự
Bài 2 : Thay zô có j kó đâu ==
(x-y)^2-4
x^2-16(x+y)^2
8x^3+36x^2y+5xy^2+27y^3
rut gon
a)(x^2+4)(x^2-4)-(x^2+1)(x^2-1)
b)(y-3)(y+3)(y^2+9)(y^2+2)(y^2-2)
Bài 1
( x - y )2 - 4
= ( x - y )2 - 22
= ( x - y - 2 )( x - y + 2 )
x2 - 16( x + y )2
= x2 - 42( x + y )2
= x2 - ( 4x + 4y )2
= ( x2 - 4x - 4y )( x2 - 4x + 4y )
8x3 + 36x2y + 54xy2 + 27y3
= ( 2x )3 + 3.( 2x )2.3y + 3.2x.( 3y )2 + ( 3y )3
= ( 2x + 3y )3
Bài 2.
a) ( x2 + 4 )( x2 - 4 ) - ( x2 + 1 )( x2 - 1 )
= [ ( x2 )2 - 42 ] - [ ( x2 )2 - 12 ]
= x4 - 16 - x4 + 1
= -15
b) ( y - 3 )( y + 3 )( y2 + 9 )( y2 + 2 )( y2 - 2 )
= [ ( y - 3 )( y + 3 )( y2 + 9 ) ][ ( y2 + 2 )( y2 - 2 ) ]
= { [ ( y - 3 )( y + 3 ) ]( y2 + 9 ) }[ ( y2 )2 - 22 ]
= [ ( y2 - 9 )( y2 + 9 ) ]( y4 - 4 )
= ( y4 - 81 )( y4 - 4 )
= y4( y4 - 4 ) - 81( y4 - 4 )
= y8 - 4y4 - 81y4 + 324
= y8 - 85y4 + 324
Tìm nghiệm nguyên của các phương trình:
1) \(x^2y^2-2xy=x^2+16y^2\)
2) \(3x^2y^2+x^2+y^2=5xy\)