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11 tháng 10 2020

Ta có: \(3x^2\left(y-x\right)+6x^2\left(x-y\right)^2\)

\(=3x^2\left(y-x\right)+6x^2\left(y-x\right)^2\)

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

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

11 tháng 10 2020

3x2( y - x ) + 6x2( x - y )2

= 3x2( y - x ) + 6x2( y - x )2

= 3x2( y - x )[ 1 + 2( y - x ) ]

= 3x2( y - x )( 2y - 2x + 1 )

10 tháng 10 2020

a) \(3^2\left(y-x\right)+6x^2\left(x-y\right)^2\)

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

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

b) \(x^4-3x^3+3x-1\)

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

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

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

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

14 tháng 10 2021

a: \(x^2-2xy+y^2+3x-3y-4\)

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

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

 

 

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: \(x^2-y^2-2x+2y\)

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

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

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

9 tháng 10 2016

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

=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)2

=(2x-3y)(x-y-2y+3x)-(4x-3y)2

=(2x-3y)(4x-3y)-(4x-3y)2

=(4x-3y)(2x-3y-4x+3y)

=(4x-3y))(-2x)

6 tháng 1 2018

Ta có (6x+5)2(3x+2)(x+1)-35

= (36x2+60x+25)(3x2+5x+2)-35 (1)

Đặt a=3x2+5x+2

=> 12a+1= 12(3x2+5x+2)+1 =36x2+60x+25

Thay a=3x2+5x+2 vào (1) ta được

(12a+1).a-35=12a2+a-35

= 12a2-20a+21a-35

= 4a(3a-5)+7(3a-5)

= (3a-5)(4a+7) (2)

Thay 3x2+5x+2=a vào (2) ta được

(9x2+15x+6-5)(12x2+20x+8+7)

= (9x2+15x+1)(12x2+20x+15)

Ta có: \(\left(6x+5\right)^2\left(3x+2\right)\left(x+1\right)-35\)

\(=\left(36x^2+60x+25\right)\left(3x^2+5x+2\right)-35\)(1)

Đặt \(3x^2+5x+2=y\)

\(\left(1\right)=\left(12y+1\right)y-35\)

\(=12y^2+y-35\)

\(=\left(3y-5\right)\left(4y+7\right)\)

\(=\left(9x^2+15x+1\right)\left(12x^2+20x+15\right)\)

10 tháng 1 2023

Bài `1:`

`a)3x^3+6x^2=3x^2(x+2)`

`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`

Bài `2:`

`a)(2x-1)^2-25=0`

`<=>(2x-1-5)(2x-1+5)=0`

`<=>(2x-6)(2x+4)=0`

`<=>[(x=3),(x=-2):}`

`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`

`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`

`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`

`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`

15 tháng 9 2021

\(A=4x^2+6x=2x\left(2x+3\right)\)

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

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

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

15 tháng 9 2021

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

9 tháng 8 2016

x3+y(1-3x2)+x(3y2-1)-y3

= x3-3x2y+3xy2-y3+y-x

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

=(x-y)(x2-2xy+y2-1)

9 tháng 8 2016

cái chỗ kia giải thích dùm mìh đy : \(x^3-3x^2y+3xy^2-y^3+y-x\)