b, \(2x\left(x-2\right)-\left(2-x\right)^2=2x\left(x-2\right)+\left(x-2\right)^2=\left(x-2\right)\left(2x+x-2\right)=\left(x-2\right)\left(3x-2\right)\)
>> T sẽ ko lm hết đâu :v <<
a) \(4x^3-14x^2\)
\(2x^2.\left(2x-7\right)\)
b) \(2x\left(x-2\right)-\left(2-x\right)^2\)
\(2x\left(x-2\right)-\left(-x\left(x-2\right)\right)^2\)
\(2x\left(x-2\right)-\left(x-2\right)^2\)
\(\left(x-2\right)\left(2x-\left(x-2\right)\right)\)
\(\left(x-2\right)\left(2x-x+2\right)\)
\(\left(x-2\right)\left(x+2\right)\)
c) \(\left(x-3\right)^3+3x\)
\(x^3-9x^2+27x-27+3x\)
\(x^3-9x^2+30x-27\)
d, \(x^3-x=x\left(x^2-1\right)=x\left(x+1\right)\left(x-1\right)\)
e, \(x^3-3x^2+3x-9=\left(x^3-3x^2+3x-1\right)-8=\left(x-1\right)^3-2^3=\left(x-1-2\right)\left[\left(x-1\right)^2+2\left(x-1\right)+2^2\right]=\left(x-3\right)\left(x^2+3\right)\)
