Question

Tìm x:    \(\left(x-2\right)^3-2\left(x+1\right)^2=\left(x+1\right)^3-3\left(2-x\right)^2\)

Answer 1

phương trình <=> \(x^3-6x^2+12x-8-2\left(x^2+2x+1\right)=x^3+3x^2+3x+1-3\left(4+x^2-4x\right)\)

<=> \(x^3-x^3-6x^2-2x^2+3x^2-3x^2+12x-4x-3x-12x-8-2-1+12=0\)

bạn cộng trừ rồi nhóm lại là ra .. ^^

Answer 2

\(\left(x-2\right)^2-2\left(x+1\right)^2=\left(x+1\right)^3-3\left(2-x\right)^2\)

\(< =>x^3-3x^2.2+3.x.2^2-2^3-2\left(x^2+2x+1\right)=x^3+3.x^2.1+3.x.1^2+1^3\)\(-3\left(2^2-4x+x^2\right)\)

\(< =>x^3-6x^2+12x-8-2x^2-4x-2=x^3+3x^2+3x+1-3.2^2+3.4x-3x^2\)

\(< =>x^3-6x^2+12x-8-2x^2-4x-2-x^3-3x^2-3x-1+12-12x+3x^2=0\)

\(< =>-8x^2-7x+1=0< =>-\left(8x^2+7x-1\right)=0< =>8x^2+7x-1=0\)

\(< =>8x^2+8x-x-1=0< =>8x\left(x+1\right)-\left(x+1\right)=0< =>\left(8x-1\right)\left(x+1\right)=0\)

<=>8x-1=0 hoặc x+1=0

<=>x=1/8 hoặc x=-1