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Phương Trần Lê
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ILoveMath
26 tháng 12 2021 lúc 16:06

tách nhỏ câu hỏi ra bạn

Kudo Shinichi
26 tháng 12 2021 lúc 16:08

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

 

Lê Hoàng Thùy Linh
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Kiều Vũ Linh
20 tháng 10 2023 lúc 8:19

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

= x(x - 2y - 3)(x - 2y + 3)

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

Nguyễn Xuân Thành
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Nguyễn Lê Phước Thịnh
22 tháng 12 2023 lúc 13:13

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

Thịnh Nguyễn
Xem chi tiết
ILoveMath
25 tháng 8 2021 lúc 8:27

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

nghia
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Kiều Vũ Linh
10 tháng 12 2023 lúc 13:11

a) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 8)

= (x + 2)(x + 4)

b) 3x² - 2(x - y)² - 3y²

= (3x² - 3y²) - 2(x - y)²

= 3(x² - y²) - 2(x - y)²

= 3(x + y)(x - y) - 2(x - y)²

= (x - y)[3(x + y) - 2(x - y)]

= (x - y)(3x + 3y - 2x + 2y)

= (x - y)(x + 5y)

c) 4x² - 9y² + 4x - 6y

= (4x² - 9y²) + (4x - 6y)

= (2x - 3y)(2x + 3y) + 2(2x - 3y)

= (2x - 3y)(2x + 3y + 2)

d) x(x + 1)² + x(x - 5) - 5(x + 1)²

= [x(x + 1)² - 5(x + 1)²] + x(x - 5)

= (x + 1)²(x - 5) + x(x - 5)

= (x - 5)[(x + 1)² + x]

= (x - 5)(x² + 2x + 1 + x)

= (x - 5)(x² + 3x + 1)

e) 2xy - x² + 3y² - 4y + 1

= -x² + 2xy - y² + 4y² - 4y + 1

= -(x² - 2xy + y²) + (4y² - 4y + 1)

= -(x - y)² + (2y - 1)²

= (2y - 1)² - (x - y)²

= (2y - 1 - x + y)(2y - 1 + x - y)

= (3y - x - 1)(x + y - 1)

f) 4x¹⁶ + 81

= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9

= (2x⁸ + 9)² - 36x⁸

= (2x⁸ + 9) - (6x⁴)²

= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)

= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)

phạm thị thùy
Xem chi tiết
Trần Thanh Phương
4 tháng 11 2018 lúc 14:48

a) \(2x^2-2y^2\)

\(=2\left(x^2-y^2\right)\)

\(=2\left(x-y\right)\left(x+y\right)\)

b) \(x^2-4x+4\)

\(=x^2-2\cdot x\cdot2+2^2\)

\(=\left(x-2\right)^2\)

c) \(x^2+2x+1-y^2\)

\(=\left(x+1\right)^2-y^2\)

\(=\left(x-y+1\right)\left(x+y+1\right)\)

d) \(x^2-4x\)

\(=x\left(x-4\right)\)

e) \(x^2+10x+25\)

\(=x^2+2\cdot x\cdot5+5^2\)

\(=\left(x+5\right)^2\)

g) \(x^2-2xy+y^2-9\)

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

\(=\left(x-y-3\right)\left(x-y+3\right)\)

h) \(2x^2-2\)

\(=2\left(x^2-1\right)\)

\(=2\left(x-1\right)\left(x+1\right)\)

i) \(5x^2-5xy+9x-9y\)

\(=5x\left(x-y\right)+9\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+9\right)\)

k) \(y^2-4y+4-x^2\)

\(=\left(y-2\right)^2-x^2\)

\(=\left(y-x-2\right)\left(y+x-2\right)\)

l) \(x^2-16\)

\(=x^2-4^2\)

\(=\left(x-4\right)\left(x+4\right)\)

m) \(3x^2-3xy+2x-2y\)

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

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

o) \(3x^4-6x^3+3x^2\)

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

\(=3x^2\left(x-1\right)^2\)

Phương Thảo
4 tháng 11 2018 lúc 14:57

a) 2x2 - 2y2

 = (2x - 2y)(2x + 2y)

 = 4(x - y)(x + y)

b) x2 - 4x + 4

 = (x - 2)2

c) x+ 2x + 1 - y2

 = (x + 1)2 - y2

 = (x + 1 - y)(x + 1 + y)

d) x2 - 4x 

 = x(x - 4)

e) x+10x + 25

 = (x + 5)2

g) x2 - 2xy + y2 - 9

= (x - y)2 - 32

 = (x - y - 3)(x - y + 3)

h) 2x2 - 2

= 2(x2 - 1) 

 = 2(x - 1)(x + 1)

i) 5x- 5xy + 9x - 9y

  = 5x(x - y) + 9(x- y)

 = (5x + 9)(x - y)

k) y2 - 4y + 4 - x2

 = (y - 2)2 - x2

 = (y - 2 - x)(y - 2 + x)

l) x- 16

 = x- 42

 = (x - 4)(x + 4)

m) 3x2 - 3xy + 2x -2y

 = 3x(x - y) +2(x-y)

 = (3x + 2)(x - y)

o) 3x- 6x+ 3x2

 = 3x4 - 3x3 - 3x3 + 3x2

 = 3x3(x - 1) - 3x2(x - 1)

 = (3x- 3x2)(x - 1)

 = 3x2(x - 1)(x - 1)

 = 3x2.(x - 1)2

Nhi huỳnh
Xem chi tiết
Tư Linh
20 tháng 8 2021 lúc 7:31

a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

c) \(x^3-x^2+2\)

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

e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)

f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)

g) \(6x^2-12x=6x\left(x-2\right)\)

h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)

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

Tư Linh
20 tháng 8 2021 lúc 7:37

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

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

Nguyễn Ngọc Mai Anh
Xem chi tiết
Nguyễn Ngọc Vy
8 tháng 6 2017 lúc 21:06

a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)

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

c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)

d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)

e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)

f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)

lê thị thu huyền
8 tháng 6 2017 lúc 21:04

a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)

\(=x^2+2xy+y^2-x^2+y^2\)

\(=2y^2+2xy\)

\(=2y\left(x+y\right)\)

c) \(3x^2+6xy+3y^2-3z^2\)

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

\(=3\left[\left(x+y\right)^2-z^2\right]\)

\(=3\left(x+y+z\right)\left(x+y-z\right)\)

d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)

\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)

\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)

\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)

\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)

\(=\left(4x^2-1\right)\left(y^2-1\right)\)

Đường Quỳnh Giang
30 tháng 9 2018 lúc 5:59

\(\left(x+y\right)^2-\left(x^2-y^2\right)\)

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

\(=\left(x+y\right)\left(x+y-x+y\right)\)

\(=2y\left(x+y\right)\)

hk tốt

^^

Hiền
Xem chi tiết
l҉o҉n҉g҉ d҉z҉
29 tháng 8 2017 lúc 15:50

9x2 + 6xy + y2

= (3x)2 + 2.3x.y + y2

= (3x + y)2

b) 6x - 9 - x2 

= -(x2 - 6x + 9)

= -(x - 3)2

Hiền
29 tháng 8 2017 lúc 15:55

mọi người nếu như có thể thì ghi cả cách lm ra hộ em vs ạ