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20 tháng 4 2020
a)
\(x^3-5x^2+6x\\ \Leftrightarrow x\cdot\left(x^2-5x+6\right)\\ \Leftrightarrow x\cdot\left(x^2-2x-3x+6\right)\\ \Leftrightarrow x\cdot\left[x\cdot\left(x-2\right)-3\cdot\left(x-2\right)\right]\\ \Leftrightarrow x\cdot\left(x-3\right)\cdot\left(x-2\right)\)
b)
\(x^2-3xy+2y^2\\ \Leftrightarrow x^2-xy-2xy+2y^2\\ \Leftrightarrow x\cdot\left(x-y\right)-2y\cdot\left(x-y\right)\\ \Leftrightarrow\left(x-2y\right)\cdot\left(x-y\right)\)
c)
\(-4x^2+10x-4\\ \Leftrightarrow-2\cdot\left(2x^2-5x+2\right)\\ \Leftrightarrow-2\cdot\left(2x^2-x-4x+2\right)\\ \Leftrightarrow-2\cdot\left[x\cdot\left(2x-1\right)-2\cdot\left(2x-1\right)\right]\\ \Leftrightarrow-2\cdot\left(x-2\right)\cdot\left(2x-1\right)\)
d)
\(x^3+2x^2y-xy^2-2y^3\\ \Leftrightarrow x^2\cdot\left(x+2y\right)-y^2\cdot\left(x+2y\right)\\ \Leftrightarrow\left(x+2y\right)\cdot\left(x^2-y^2\right)\\ \Leftrightarrow\left(x+2y\right)\cdot\left(x+y\right)\cdot\left(x-y\right)\)
CB
10 tháng 2 2017
a) 5x2 - 10x = 5x( x - 2 )
b) x2 - y2 - 2x + 2y = (x2 - y2) - (2x - 2y)
= (x - y ) ( x + y)-2 (x-y)
= ( x - y) ( x + y - 2)
c) 4x2 - 4xy - 8y2 = (4x2 - 4xy + 8y2) - 9y2
= (2x - 9y2) - 3y2
= (2x - y - 3y) (2x - y + 3y)
= (2x - 4y) (2x + 2y)
= 4(x - 2y) (x + y)
CB
10 tháng 2 2017
a) 5x2 - 10x = 5x( x - 2 )
b) x2 - y2 - 2x + 2y = (x2 - y2) - (2x - 2y)
= (x - y ) ( x + y)-2 (x-y)
= ( x - y) ( x + y - 2)
c) 4x2 - 4xy - 8y2 = (4x2 - 4xy + 8y2) - 9y2
= (2x - 9y2) - 3y2
= (2x - y - 3y) (2x - y + 3y)
= (2x - 4y) (2x + 2y)
= 4(x - 2y) (x + y)
28 tháng 1 2020
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
C
0
KB
10 tháng 9 2018
1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)
2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)
3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)
4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
5 ) \(5x^2y+5xy^2+a^2x+a^2y\)
\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)
\(=\left(5xy+a^2\right)\left(x+y\right)\)
6 ) \(2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(2x+5\right)\left(x-3y\right)\)
7 ) \(ax^2-3axy+bx-3by\)
\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)
\(=x\left(ax+b\right)-3y\left(ax+b\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
8 ) \(x^2+4x-5x-20=0\)
\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
9 ) \(x^2+10x-2x-20=0\)
\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
10 ) \(x^2-6x-4x+24=0\)
\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
:D
13 tháng 6 2019
a) \(3y^2\left(2y-1\right)+y-y\left(1-y+y^2\right)-y^2+y \)
= \(6y^3-3y^2+y-y+y^2-y^3-y^2+y\)
= \(5y^3-3y^2+y\)
b)\(25x-4\left(3x-1\right)+\left(5-2x\right)7\)
= \(25x-12x+4+35-14x\)
= \(-x+39\)
c) \(11x-2\left(10x-1\right)-\left(4x-1\right)\left(-2\right)\)
= \(11x-\left(20x-2\right)-\left(-8x+2\right)\)
= \(11x-20x+2+8x-2\)
= \(-x\)
d) \(\left(\frac{1}{2x}\right)3-x\left(1-2x-\frac{1}{8x^2}\right)-x\left(x+\frac{1}{2}\right)\)
= \(\frac{3}{2x}-x+2x^2+\frac{x}{8x^2}-x^2-\frac{x}{2}\)
= \(\left(\frac{3}{2x}+\frac{1}{8x}-\frac{x}{2}\right)+x^2-x\)
= \(\left(\frac{12+1-4x^2}{8x}\right)+x^2-x\)
= \(\frac{13-4x^2}{8x}+\frac{8x^3}{8x}-\frac{8x^2}{8x}\)
= \(\frac{13-4x^2+8x^3-8x^2}{8x}\)
= \(\frac{8x^3-12x^2+13}{8x}\)
= x2 - \(\frac{3}{2}\)+\(\frac{13}{8x}\)
e) \(12\left(2-3x\right)+35x-\left(x+1\right)\left(-5\right)\)
= \(24-36x+35x-\left(-5x-5\right)\)
= \(24-36x+35x+5x+5\)
= 4x + 29
26 tháng 8 2018
Bài 1:
a, 2x(x - 3y )
= 2x2 - 6y
b, ( x - 2y ) ( x2 - 2xy + y2 )
= x3 - 2x2y + xy2 - 2x2y + 4xy2 - 2y3
= x3 -4 x2y + 5xy3 - 2y3
Bài 2:
5x( 4x2 - 2x + 1 ) - 2x ( 10x2 - 5x - 2 )
= 20x3 - 10x2 + 5x - 20x3 + 10x2 + 4x
= 9x
( don't k ❤EXO❤ )
3 tháng 9 2016
1 ) Thực hiện phép tính :
a ) \(-\frac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
b ) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2-x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2x-2-x^3\)
c ) \(\left(x^3+5x^2-2x+1\right)\left(x-7\right)\)
\(=x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\)
\(=x^4-2x^3-37x^2+15x-7\)
d ) \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2+y^3\)
e ) \(\left[\left(x^2-2xy+2y^2\right)\left(x+2y\right)-\left(x^2-4y^2\right)\left(x-y\right)\right]2xy\)
( để xem lại )
2 Tìm x
a ) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=3\)
b ) Sai đề
c ) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
( Để xem lại )
HC
5 tháng 9 2016
mình chép đúng theo đề cô cho mà sao lại sai được ,hay cô cho sai đề
\(a,=5x^2\left(x-2y^2\right)\\ b,=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\)
a) \(=5x^2\left(x-2y^2\right)\)
b) \(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\)