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AD
0
13 tháng 9 2017
a,\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)
\(=\left(x-1\right)\left(x\left(x-3\right)+8\left(x-3\right)\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
CW
29 tháng 12 2016
\(3x^2-5x+2=3x^2-3x-2x+2=3x\left(x-1\right)-2\left(x-1\right)=\left(3x-2\right)\left(x-1\right)\)
29 tháng 12 2016
b.)x^4+5x^3+15x-9
=x^4-9+5x^3+15x
=(x^2-3)(x^2+3)+5x(x^2+3)
=(x^2+3)(x^2-3+5x)
NH
1
5 tháng 8 2017
\(x^3+4x^2-29x+24\)
\(=x^3-4x^2+3x+8x^2-32x+24\)
\(=x\left(x^2-4x+3\right)+8\left(x^2-4x+3\right)\)
\(=\left(x^2-4x+3\right)\left(x+8\right)\)
\(=\left(x^2-3x-x+3\right)\left(x+8\right)\)
\(=\left[x\left(x-3\right)-\left(x-3\right)\right]\left(x+8\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
29 tháng 8 2020
a) 4x3 + 4x2 + 6x
= 2x( 2x2 + 2x + 3 )
b) x3 + 4x2 - 29x + 24
= x3 + 8x2 - 4x2 - 32x + 3x + 24
= ( x3 - 4x2 + 3x ) + ( 8x2 - 32x + 24 )
= x( x2 - 4x + 3 ) + 8( x2 - 4x + 3 )
= ( x2 - 4x + 3 )( x + 8 )
= ( x2 - x - 3x + 3 )( x + 8 )
= [ x( x - 1 ) - 3( x - 1 ) ]( x + 8 )
= ( x - 1 )( x - 3 )( x + 8 )
c) x3 + 21x2 + 134x + 240
= x3 + 10x2 + 11x2 + 110x + 24x + 240
= ( x3 + 11x2 + 24x ) + ( 10x2 + 110x + 240 )
= x( x2 + 11x + 24 ) + 10( x2 + 11x + 24 )
= ( x + 10 )( x2 + 11x + 24 )
= ( x + 10 )( x2 + 3x + 8x + 24 )
= ( x + 10 )[ x( x + 3 ) + 8( x + 3 ) ]
= ( x + 10 )( x + 8 )( x + 3 )
d) 25x3 - 25x2y - x + y
= 25x2( x - y ) - 1( x - y )
= ( x - y )( 25x2 - 1 )
= ( x - y )( 5x - 1 )( 5x + 1 )
e) 2x5y - 4x3y + 2xy
= 2xy( x4 - 2x2 + 1 )
Đặt t = x2
= 2xy( t2 - 2t + 1 )
= 2xy( t - 1 )2
= 2xy( x2 - 1 )2
= 2xy[ ( x - 1 )( x + 1 ) ]2
= 2xy( x - 1 )2( x + 1 )2
g) x3 - 3x + 2
= x3 + 2x2 - 2x2 - 4x + x + 2
= ( x3 - 2x2 + x ) + ( 2x2 - 4x + 2 )
= x( x2 - 2x + 1 ) + 2( x2 - 2x + 1 )
= ( x + 2 )( x2 - 2x + 1 )
= ( x + 2 )( x - 1 )2
GH
13 tháng 10 2017
c) 2x2 + 10x + 8
= 2x2 + 2x + 8x + 8
= 2x( x + 1) + 8(x + 1)
= 2(x + 1)(x + 4)
d) x2 - 7xy + 10y2
= x2 - 2xy - 5xy + 10y2
= x(x - 2y) - 5y(x - 2y)
= (x - 2y)(x - 5y)
e) x4 + 4x2 - 5
= x4 - x2 + 5x2 - 5
= x2(x2 - 1) + 5(x2 - 1)
= (x - 1)(x + 1)(x2 + 5)
f) x3 - 7x - 6
= x3 - x - 6x - 6
= x(x2 - 1) - 6(x + 1)
= x(x - 1)(x + 1) - 6(x + 1)
= (x + 1)(x - 1)(x - 6)
pn coi kt lại nhé
29 tháng 10 2018
a)\(\left(3x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(3x-\dfrac{1}{3}\right)\left(9x^2+x+\dfrac{1}{9}\right)\)
b)\(x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x^2+5x-24\right)\left(x-1\right)\)
\(=\left(x^2+8x-3x-24\right)\left(x-1\right)\)
\(=\left[x\left(x+8\right)-3\left(x+8\right)\right]\left(x-1\right)\)
\(=\left(x-3\right)\left(x+8\right)\left(x-1\right)\)
c)\(x^2+7x+12=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
d)\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
2 tháng 2 2017
1, (x2-x+2)2-(x-2)2=(x2-x+2-x+2)(x2-x+2+x-2)=(x2-2x+4)x2
2,a.x3+4x2-29x+24=0
\(\Leftrightarrow\)x3-3x2+7x2-21x-8x+24=0
\(\Leftrightarrow\)(x3-3x2)+(7x2-21x)-(8x+24)=0
\(\Leftrightarrow\)x2(x-3)+7x(x-3)-8(x-3)=0
\(\Leftrightarrow\)(x-3)(x2-x+8x-8)=0
\(\Leftrightarrow\)(x-3)(x-1)(x+8)=0
\(\Leftrightarrow\)\(\left[\begin{matrix}x-3=0\\x-1=0\\x+8=0\end{matrix}\right.\)\(\left[\begin{matrix}x=3\\x=1\\x=-8\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-8;1;3\right\}\)
b. đặt x2-x=y ta có:
y2-14y+24=0 \(\Leftrightarrow\)(y2-2.7y+49)-25=0 \(\Leftrightarrow\)(y-7)2-52=0 \(\Leftrightarrow\)(y-12)(y-2)=0 \(\Leftrightarrow\left[\begin{matrix}y=12\\y=2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}x^2-x=12\\x^2-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2-x-12=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\left(x+3\right)\left(x-4\right)=0\\\left(x-2\right)\left(x+1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-3\\x=4\\x=2\\x=-1\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-3;-1;2;4\right\}\)
3 tháng 2 2017
3.ta có : 5x2+5y2+8xy+2x-2y+2=0
\(\Leftrightarrow\)(4x2+8xy+4y2)+(x2+2x+1)+(y2-2y+1)=0
\(\Leftrightarrow\)(2x+2y)2+(x+1)2+(y-1)2=0
lại có (2x+2y)2+(x+1)2+(y-1)2\(\ge\)0 dấu = chỉ sảy ra khi và chỉ khi \(\left\{\begin{matrix}\left(2x+2y\right)^2=0\\\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2x+2y=0\\x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
vậy x=-1 và y=1
\(=x^3+8x^2-4x^2-32x+3x+24\\ =\left(x+8\right)\left(x^2-4x+3\right)=\left(x+8\right)\left(x^2-x-3x+3\right)\\ =\left(x+8\right)\left(x-1\right)\left(x-3\right)\)