(3x-1)(y-3)=30
x-2xy+4y=7
3x-6xy-8y=10
2x-5xy+10y=9
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 11 2016
3(x2-2xy+y2):10(x-y)
3(x-y)2:10(x-y)
3(x-y):10
5(x2-xy+y2):(x+y)(x2-xy+y2)
5:x+y
ở nơi 5y thiếu ^2 nhé
A
1 tháng 10 2020
a, \(5ax-15ay+20a=5a\left(x-5y+4\right)\)
b, sai
c, \(3ab\left(x+y\right)+3a\left(y-x\right)=3ab\left(x+y\right)-3a\left(x+y\right)=\left(3ab-3a\right)\left(x+y\right)\)
d, \(x^2-xy+2x-2y=x\left(x+2\right)-y\left(x+2\right)=\left(x-y\right)\left(x+2\right)\)
Tượng tự ...
1 tháng 10 2020
a) 5ax - 15ay + 20a = 5a( x - 3y + 4 )
b) 6xy - 12x - 8y = 2( xy - 6x - 4y )
c) 3ab( x - y ) + 3a( y - x ) = 3ab( x - y ) - 3a( x - y ) = ( x - y )( 3ab - 3a ) = 3a( x - y )( b - 1 )
d) x2 - xy + 2x - 2y = x( x - y ) + 2( x - y ) = ( x - y )( x + 2 )
e) ax2 - 5x2 - ax + 5x + a - 5 = x2( a - 5 ) - x( a - 5 ) + ( a - 5 ) = ( a - 5 )( x2 - x + 1 )
g) x2y - 4xy2 + 4y3 - 36yz2 = y( x2 - 4xy + 4y2 - 36z2 ) = y[ ( x2 - 4xy + 4y2 ) - 36z2 ] = y[ ( x - 2y )2 - ( 6z )2 ] = y( x - 2y - 6z )( x - 2y + 6z )
h) 4xy - x2 - 4y2 + m2 - 6m + 9
= ( m2 - 6x + 9 ) - ( x2 - 4xy + 4y2 )
= ( m - 3 )2 - ( x - 2y )2
= ( m - 3 - x + 2y )( m - 3 + x - 2y )
i) x2 + x - 12 = x3 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
k) 5x2 + 14x - 3 = 5x2 - x + 15x - 3 = x( 5x - 1 ) + 3( 5x - 1 ) = ( 5x - 1 )( x + 3 )
m) x2 - 5xy + 4y2 = x2 - xy - 4xy + 4y2 = x( x - y ) - 4y( x - y ) = ( x - y )( x - 4y ) < đã sửa đề >
n) 3x2 - 5xy + 2y2 + 4x - 4y = ( 3x2 - 5xy + 2y2 ) + ( 4x - 4y ) = ( 3x2 - 3xy - 2xy + 2y2 ) + 4( x - y ) = [ 3x( x - y ) - 2y( x - y ) ] + 4( x - y ) = ( x - y )( 3x - 2y ) + 4( x - y ) = ( x - y )( 3x - 2y + 4 )
f) 2x3 + 4x2y + 2xy2 = 2x( x2 + 2xy + y2 ) = 2x( x + y )2
NC
1 tháng 10 2020
a) \(5ax-15ay+20a\)
\(=5a\left(x-3y+4\right)\)
b) \(6xy-12x-8y\)
\(=6\left(xy-2x-3y\right)\)
c) \(3ab\left(x-y\right)+3a\left(y-x\right)\)
\(=3a\left(x-y\right)\left(b-1\right)\)
d) \(x^2-xy+2x-2y\)
\(=\left(x+2\right)\left(x-y\right)\)
NC
1 tháng 10 2020
e) \(ax^2-5x^2-ax+5x+a-5\)
\(=\left(a-5\right)\left(x^2-x+1\right)\)
TA
3
11 tháng 8 2023
`a,x^3 - 3x^2 + 1 - 3x`
`=x^3 + 1 - 3x^2 - 3x`
`=(x^3 + 1) - 3x(x+1)`
`=(x+1)(x^2 - x + 1) - 3x(x+1)`
`=(x+1)(x^2 - x + 1 - 3x)`
`=(x+1)(x^2 - 4x + 1)`
`b,x^2 + 4x - 2xy - 4y + y^2`
`=(x^2 -2xy + y^2) + (4x-4y)`
`=(x-y)^2 + 4(x-y)`
`=(x-y)(x-y+4)`
`c,3x^2 -6xy + 3y^2 - 12z^2`
`=3(x^2 -2xy +y^2 - 4z^2)`
`=3[(x-y)^2 - (2z)^2]`
`=3(x-y-2z)(x-y+2z)`
11 tháng 8 2023
a: =x^3+1-3x^2-3x
=(x+1)(x^2-x+1)-3x(x+1)
=(x+1)(x^2-x+1-3x)
=(x+1)(x^2-4x+1)
b: =x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
c: =3(x^2-2xy+y^2-4z^2)
=3[(x-y)^2-4z^2]
=3(x-y-2z)(x-y+2z)
AH
Akai Haruma
Giáo viên
29 tháng 5 2023
a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
ND
0
27 tháng 9 2021
a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)
\(=x^2-2xy+y^2+4y^2+4y+1+50\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)
27 tháng 9 2021
a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)
\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)
c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)
\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)
d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)
\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)
\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)
L
26 tháng 10 2019
\(a)xy+3x-2y=11\)
\(\Leftrightarrow xy+3x-2y-6=5\)
\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)
L
26 tháng 10 2019
\(b)2x^2-2xy+x-y=12\)
\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)
\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)
\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)
\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
Vì 2x+1 luôn lẻ
\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)