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a) \(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
c) x^2y+xy+x+1=xy(x+1)+x+1=(x+1)(xy+1) d)x^2-ax-bx+ab=x(x-a)-b(x-a)=(x-a)(x-b) d) (x^2-4y^2)-(2x+4y)=(x+2y)(x-2y)-2(x+2y)=(x+2y)(x-2y-2)
x2 - x - y2 - y
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
***
9x2 + y2 - 16z2 + 6xy
= (3x + y)2 - (4z)2
= (3x + y - 4z)(3x + y + 4z)
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a3 - a2x - ay + xy
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
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2x2 - 8y2 + 3x + 6y
= 2(x2 - 4y2) + 3(x + 2y)
= 2(x - 2y)(x + 2y) + 3(x + 2y)
= (x + 2y)(2x - 4y + 3)
***
xy(x + y) + yz(y + z) + xz(x + z) + 2xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + z)
= y(x + y + z)(x + z) + xz(x + z)
= (x + z)(xy + y2 + yz + xz)
= (x + z)[y(x + y) + z(x + y)]
= (x + z)(x + y)(y + z)
\(10\left(x-y\right)-8y\left(y-x\right)\)
\(=10\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10+8y\right)\)
\(=2\left(x-y\right)\left(5+4y\right)\)
a) 10(x-y)-8y(y-x)= 10(x-y)+8y(x-y) = (x-y)(10+8y)=2(x-y)(5+4y)
b) Bạn xem lại đầu bài nhé !
\(x-y-a\left(x-y\right)\)
\(=\left(x-y\right)-a\left(x-y\right)\)
\(=\left(x-y\right)\left(1-a\right)\)
\(x-y-a\left(x-y\right)\)
\(=\left(x-y\right)-a\left(x-y\right)\)
\(=\left(1-a\right).\left(x-y\right)\)
\(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a-x\right)\left(a^2-y\right)\)
\(4x^2-y^2+4x+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2=\left(2x-y+1\right)\left(2x+y+1\right)\)
\(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
a)a3 - a2x - ay +xy
=(a3 - a2x) - (ay - xy)
=a2(a-x) - y(a-x)
=(a-x).(a2 - y)
a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)
\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)
c) \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)
\(=[\left(3x+2\right)-\left(x-1\right)]^2=\left(3x+2-x+1\right)^2=\left(2x+3\right)^2=\left(2x+3\right)\left(2x+3\right)\)
x( a - b ) - a + b = x( a - b ) - ( a - b ) = ( a - b )( x - 1 )
a( x - y ) - x + y = a( x - y ) - ( x - y ) = ( x - y )( a - 1 )
mấy ý kia không sửa được nữa nên nghỉ:)