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)

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

bạn nhân ra ra được mà chẳng hạn x(x+1)=x²+x

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: \(x^2-y^2-2x+2y\)

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

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

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

a: \(x^2-2xy+y^2+3x-3y-4\)

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

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

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

`a)2x(x-3)-x+3=0`

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

`<=>(x-3)(2x-1)=0`

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

`<=>(x+2)(x+3)=0`

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

Đúng hết mà?

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

=(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)

tách nhỏ câu hỏi ra bạn

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

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

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

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

a)(x+y)²-(x-y)²

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

=2y.2x=4xy

b)(3x+1)²-(x+1)²

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

=2x.(4x+2)

=4x(2x+1)

c) x³+y³+z³-3xyz

= (x+y)³- 3xy(x+y) +z³-3xyz

=(x+y+z)( x²+2xy+y²-xz-yz+z²)-3xy(x+y+z)

=(x+y+z)(x²+y²+z²-xy-xz-yz)

Phân tích đa thức sau thành nhân tử :

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

b) \(x^3+y^3+z^3-3xyz\)

\(x^3+y^3+z^3-3xyz\)

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

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)