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23 tháng 9 2017

a) x3-2x2-x+2

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

=x(x2-1)-2(x2-1)

=(x2-1)(x-2)

b)

x2+6x-y2+9

=x2+6x+9-y2

=(x+3)2-y2

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

1 tháng 11 2021

\(=5^{^2}.\left(x+5\right)^2-3^2.\left(x+7\right)^2\)

\(=\left(5x+25\right)^2-\left(3x+21\right)^2\)

\(=\left(5x+25+3x+21\right)\left(5x+25-3x-21\right)\)

\(=\left(8x+46\right)\left(2x+4\right)\)

\(=4\left(2x+23\right)\left(x+2\right)\)

1 tháng 11 2021

  = 52 ( x + 5)2 - 32 (x +7)2

=[ 5 ( x +5) ]2 - [ 3 ( x + 7) ]2

= ( 5x + 25)2 - ( 3x + 21)2

= ( 5x + 25 - 3x - 21) - ( 5x + 25 + 3x + 21)

= ( 2x +4) - ( 8x +46)

= -6x - 42

= -6 ( x + 7)

7 tháng 5 2022

`a)5(x-y)-y(x-y)`

`=(x-y)(5-y)`

`b)x^2-6x-y^2+9`

`=(x^2-6x+9)-y^2`

`=(x-3)^2-y^2`

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

25 tháng 9 2017

\(x^5-x^4-x^3-x^2-x-2=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)

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

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

3 tháng 9 2021

\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2\left(x^2-2x+1\right)-5x+5-1\right)=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)

3 tháng 9 2021

\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)

\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left[2\left(x^2-2x+1\right)-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)

\(=\left(x-1\right)\left(2x^2-9x+6\right)\)

28 tháng 9 2020

a) \(14\left(x-y\right)^2+21\left(y-x\right)\)

\(=14\left(x-y\right)^2-21\left(x-y\right)\)

\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)

\(=7\left(x-y\right)\left(2x-2y-3\right)\)

b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)

\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)

\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)

28 tháng 9 2020

c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)

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

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

\(=\left(x-3\right)^2\left(x^2+6x+9-x^2\right)\)

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

d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)

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

\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)

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

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

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

6 tháng 10 2021

a) \(-10x^3+2x^2=0\)

\(\Rightarrow-2x^2\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(5x\left(x-2016\right)-x+2016=0\)

\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)

\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)

a: Ta có: \(-10x^3+2x^2=0\)

\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)

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