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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

5 tháng 9 2021

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

 

 

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

15 tháng 10 2021

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

15 tháng 10 2021

1.

a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

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

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

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

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) \(5ax-15ay+20a\)

\(=5a\left(x-3y+4\right)\)

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) \(x^2-xy+2x-2y\)

\(=\left(x+2\right)\left(x-y\right)\)

1 tháng 10 2020

e) \(ax^2-5x^2-ax+5x+a-5\)

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

8 tháng 12 2019

Bài làm

a) xy + y2 - x - y

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

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

= ( x + y )( y - 1 )


b) 25 - x2 + 4xy - 4y2

= 25 - ( x2 - 4xy + 4y2 )

= 25 - ( x - 2y )2

= ( 5 - x + 2y )( 5 + x - 2y )

c) xy + xz - 2y - 2z

= ( xy + xz ) - ( 2y + 2z )

= x( y + z ) - 2( y + z )

= ( y + z )( x - 2 )


d) x2 - 6xy + 9y2 - 25z2

= ( x2 - 6xy + 9y2 ) - 25z2

= ( x - 3y )2 - 25z2

= ( x - 3y - 5z )( z - 3y + 5z )


e) 3x2 - 3y2 - 12x + 12y

= 3( x - y )( x + y ) - 12( x - y )

= ( x - y )[ 3( x + y ) - 12 ]

f) 4x3 + 4xy2 + 8x2y - 16x

= 4x( x2 + y2 + 2xy - 4 )

= 4x[ ( x + y)2 - 4 ]

= 4x( x + y - 2 )( x + y + 2 )


g) x2 - 5x + 4

= x2 - x - 4x + 4

= x( x - 1 ) - 4( x - 1 )

= ( x - 1 )( x - 4 )


h) x4 + 5x2 + 4

= x4 + x2 + 4x2 + 4

= x2( x2 + 1 ) + 4( x2 + 1 )

= ( x2 + 1 )( x2 + 4 )


i) 2x2 + 3x - 5

= 2x2 - 5x + 2x - 5

= 2x( x + 1 ) - 5( x + 1 )

= ( x + 1 )( 2x - 5 )


k) x3 - 2x2 + 6x - 5 ( không biết làm )
l) x2 - 4x + 3

= ( x2 - 4x + 4 ) - 1

= ( x - 2 )2 - 1

= ( x - 3 )( x - 1 )

# Học tốt #

a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)

b: \(=25-\left(x-2y\right)^2\)

\(=\left(5-x+2y\right)\left(5+x-2y\right)\)

26 tháng 10 2021

a: \(=\left(3-x\right)\left(x+1\right)\)

b: \(=3x\left(x-y\right)-5\left(x-y\right)\)

=(x-y)(3x-5)

c: \(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

26 tháng 10 2021

a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)

b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)

d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)

e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)

f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)

g) \(=y\left(y^2-2xy+x^2-y\right)\)

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

i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)

13 tháng 8 2021

a) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)

d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)

13 tháng 8 2021

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