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a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)

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

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

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

b) Ta có: \(9-x^2+2xy-y^2\)

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

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

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

a) Ta có: \(x^2-2xy+y^2-2x+2y\)

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

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

b) Ta có: \(x^2-4x+4-x^2y+2xy\)

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

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

14 tháng 8 2021

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

b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

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

b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)

21 tháng 7 2021

36x2-12x+1-y2=(36x2-12x+1)-y2=(6x-1)2-y2=(6x-1+y)(6x-1-y)

21 tháng 7 2021

a2-2a+1-49b2=(a2-2a+1)-49b2=(a-1)2-(7b)2=(a-1-7b)(a-1+7b)

2 tháng 9 2018

\(x^3+8y^3+2xy^2+x^2y\)

\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)

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

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

a: =(x^2-x+1)(x^2+x+1)

b: =x^2-6xy+9y^2=(x-3y)^2

c: =5x(x^2-2xy+y^2)

=5x(x-y)^2

d: =(x-3)^2

e: =(2y-z)(4x+7y)

2 tháng 1 2023

a)HĐT:(x^2+1-x)(x^2+1+x)

b)=x^2-2.x.3y+(3y)^2

c)=5x(x^2-2xy+y^2)

=5x(x-y)^2

d)x^2-2.3.x+3^2

=(x-3)^2

e)(2y-z)+7y(2y-z)

=(2y-z)(1+7y)

22 tháng 12 2023

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: \(x^2-y^2-2x+2y\)

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

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

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

14 tháng 11 2021

x²-6xy+9y²-36 =(x²-6xy+9y²)-36 =(x-3y)²-6² =(x-3y+6)(x-3y-6)

20 tháng 8 2021

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

2, \(x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)

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

4, \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)

\(1,\)

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

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

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

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

\(2,\)

\(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

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

\(3,\)

\(x^2y-x^3-9y+9x\)

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

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

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

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

\(4,\)

\(x^4+2x^3+x^2\)

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

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

\(5,\)

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

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