Đặt thành thừa số chung :
a) xy + x + 8y + 8
b) x2 - 1
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a) xy + x + 8y + 8
= x.(y + 1) + 8.(y + 1)
= (y + 1).(x + 8)
b) \(x^2-x-\frac{2}{3}.x+\frac{2}{3}\)
\(=x.\left(x-1\right)-\frac{2}{3}.\left(x-1\right)\)
\(=\left(x-1\right).\left(x-\frac{2}{3}\right)\)
c) x2 - 1
= x2 + x - x - 1
= x.(x + 1) - (x + 1)
= (x + 1).(x - 1)
a) (xy+x) +(8y+8)=x(y+1)+8(y+1)=(x+8)(y+1)
b) (x2-x) -(2/3x-2/3)=x(x-1)+2/3(x-1)=(x+2/3)(x-1)
c) x2-1= (x-1)(x+1)
\(x^2-xy-xy+y^2\)
\(=\left(x^2-xy\right)-\left(xy-y^2\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(\left(x-y\right)\times\left(x-y\right)\)
\(a\left(x-1\right)+b\left(1-x\right)=a\left(x-1\right)-b\left(x-1\right)=\left(x-1\right)\left(a-b\right)\)
a) ( x + 1) + b ( 1 - x )
= ( x + 1 ) - b( x- 1 )
= ( 1 - b )( x- 1 )
****
\(x^2-1=\left(x+1\right)\left(x-1\right)\)(Nhớ k cho mình với nhé!)
h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)
i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)
x2 - 1 = x2 - x + x - 1 = x . (x - 1) + (x - 1) = (x - 1) . (x + 1)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
\(x^2-1\) = \(x^2-x+x-1\) = \(x\left(x-1\right)+\left(x-1\right)\) = \(\left(x-1\right)\left(x+1\right)\)
a, y(x+1+8+8)
y(x+17)
b, x(x-1)
a) xy + x + 8y + 8
= x ( y + 1 ) + 8 ( y + 1 )
= ( y + 1 ) . ( x + 8 )
b) x2 - 1
= x2 - x + x - 1
= x ( x - 1 ) + ( x - 1 )
= ( x - 1 ) . ( x + 1 )