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a, mình nghĩ đề là cm đẳng thức nhé
\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)
Vậy ta có đpcm
b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)
\(=-5y-9+xy=VP\)
Vậy ta có đpcm
c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)
Vậy ta có đpcm
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c) (xy-1).(xy+5)
= x2y2+5xy-xy-5
=x2y2+4xy-5
a) b) d) bạn có thể ghi rõ được ko
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bn hãy xem cách tui làm phía duoi la làm dc, cứ xem có j chung thi rút nó ra
a) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
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a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
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1) 2x + 2y - x(x+y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2/ 5x2 - 5xy -10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10(x - y)
3/ 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
1) 2x + 2y - x(x + y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2) 5x2 - 5xy - 10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10)(x - y)
= 5(x - 2)(x - y)
3) 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
4) 2x2 + 2y2 - x2z + z - y2z - 2
= 2(x2 + y2 - z(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2)
5) x2 + xy - 5x - 5y
= x(x + y) - 5(x + y)
= (x - 5)(x + y)
6) x(2x - 7) - 4x + 14
= x(2x - 7) - 2(2x - 7)
= (x - 2)(2x - 7)
7)x2 - 3x + xy - 3y
= x(x + y) - 3(x + y)
= (x - 3)(x + y)
\(5xy.\left(3x^2-xy+4\right)\)
\(=5xy.3x^2-5xy.xy+5xy.4\)
\(=15x^3y-5x^2y+20xy\)