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TP
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NV
23 tháng 9 2017 lúc 23:18

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)

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N2
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PT
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H24
13 tháng 12 2021 lúc 20:28

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

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DA
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LD
1 tháng 11 2020 lúc 21:22

( x + y + z )3 - x3 - y3 - z3

= [ ( x + y + z )3 - x3 ] - ( y3 + z3 )

= ( x + y + z - x )[ ( x + y + z )2 + ( x + y + z )x + x2 ] - ( y + z )( y2 - yz + z2 )

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

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

= ( y + z )( 3x2 + 3yz + 3zx + 3xy )

= 3( y + z )( x2 + yz + zx + xy )

= 3( y + z )[ ( x2 + zx ) + ( xy + yz ) ]

= 3( y + z )[ x( x + z ) + y( x + z ) ]

= 3( y + z )( x + z )( x + y )

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H24
1 tháng 11 2020 lúc 21:24

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

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

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

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

\(=\left(y+z\right).\left[x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-\left(y^2-yz+z^2\right)\right]\)

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

\(=\left(y+z\right).\left(3x^2+3xy+3yz+3xz\right)\)

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

\(=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]\)

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

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

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ML
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LL
6 tháng 10 2021 lúc 20:35

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) \(5x\left(x-2016\right)-x+2016=0\)

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

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

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

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NT
6 tháng 10 2021 lúc 21:30

a: Ta có: \(-10x^3+2x^2=0\)

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

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

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NV
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HH
12 tháng 10 2021 lúc 10:44

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

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NT
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H24
19 tháng 12 2023 lúc 19:54

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

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KN
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NK
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KR
14 tháng 12 2023 lúc 21:43

`#3107.101107`

`x^2 - y^2 + 10x - 10y`

`= (x^2 - y^2) + (10x - 10y)`

`= (x - y)(x + y) + 10(x - y)`

`= (x + y + 10)(x - y)`

_____

Sử dụng HĐT:

`A^2 - B^2 = (A - B)(A + B).`

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