a/ ab-b2-a+b
=a(b-1)-b(b-1)
=(b-1)(a-b)
b/ 64-(x-1)3
= (4-x+1)[16+4(x-1)+(x-1)2 ]
c/ 2x2+4x-70
=2(x2+2x-35)
a) \(ab-b^2-a+b=b\left(a-b\right)-\left(a-b\right)=\left(b-1\right)\left(a-b\right)\)
b) \(64-\left(x-1\right)^3\)
\(=8^3-\left(x-1\right)^3\\ =\left(8-x-1\right)\left[8^2+8.\left(x-1\right)+\left(x-1\right)^2\right]\\ =\left(7-x\right)\left(64+8x-8+x^2-2x+1\right)\\ =\left(7-x\right)\left(x^2+57+6x\right)\)
c) \(2x^2+4x-70\)
\(=2x^2-10x+14x-70\\ =\left(2x^2-10x\right)+\left(14x-70\right)\\ =2x\left(x-5\right)+14\left(x-5\right)\\ =\left(2x+14\right)\left(x-5\right)\\ =2\left(x+7\right)\left(x-5\right)\)
a) \(ab-b^2-a+b\)
⇔ \(\left(ab-a\right)-\left(b^2-b\right)\)
⇔ \(a\left(b-1\right)-b\left(b-1\right)\)
⇔ \(\left(a-b\right)\left(b-1\right)\)
a) Ta có: \(ab-b^2-a+b\)
\(=b\left(a-b\right)-\left(a-b\right)\)
\(=\left(a-b\right)\left(b-1\right)\)
b) Ta có: \(64-\left(x-1\right)^3\)
\(=\left(4-x+1\right)\left[16+4\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=\left(5-x\right)\left(16+4x-4+x^2-2x+1\right)\)
\(=\left(5-x\right)\left(x^2+2x+13\right)\)
c) Ta có: \(2x^2+4x-70\)
\(=2\left(x^2+2x-35\right)\)
\(=2\left(x+7\right)\left(x-5\right)\)
