a, Tính ra thôi:(x−1)3−x(x+1)2=5x(2−x)−11(x+2)(x−1)3−x(x+1)2=5x(2−x)−11(x+2)
<=> x3−3x2+3x−1−x(x2+2x+1)=10x−5x2−11x−22x3−3x2+3x−1−x(x2+2x+1)=10x−5x2−11x−22
<=>−5x2+2x−1=−x−5x2−22−5x2+2x−1=−x−5x2−22
<=>−5x2+5x2+2x+x=−22+1−5x2+5x2+2x+x=−22+1
<=> 3x=-21
=>x=-7.
b, (x−3)(x+4)−2(3x−2)=(x−4)2(x−3)(x+4)−2(3x−2)=(x−4)2
<=> x2+4x−3x−12−6x+4=x2−2.4x+42x2+4x−3x−12−6x+4=x2−2.4x+42
<=>x2−5x−8=x2−8x+16x2−5x−8=x2−8x+16
<=> x2−x2−5x+8x=16+8x2−x2−5x+8x=16+8
<=> 3x=24
=> x=8
a: \(\Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x-22\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\)
=>3x=-21
hay x=-7
b: \(\Leftrightarrow x^2+x-12-6x+4=x^2-8x+16\)
=>-8x+16=-5x-8
=>-3x=-24
hay x=8
