Question

giải pt

\(x^3+x+6=2\left(x+1\right)\sqrt{3+2x-x^2}\)

Answer 1

@Cold Wind

<=>\(x^3+x+6=2\left(x+1\right)\sqrt{3+2x-x^2}=2x\sqrt{3+2x-x^2}+2\sqrt{3+2x-x^2}\)<=>\(x^3+x+6-4x-4=2x\left[\sqrt{3+2x-x^2}-2\right]+\left[\sqrt{3+2x-x^2}-2\right]\)

<=>\(\left(x^3-x-2x+2\right)\left(\sqrt{3+2x-x^2}+2\right)=2x\left[3+2x-x^2-4\right]+\left[3+2x-x^2-4\right]\)\(\left(\sqrt{3+2x-x^2}+2\right)\left[x\left(x^2-1\right)-2\left(x-1\right)\right]=-2\left(x+1\right)\left(x-1\right)^2\)

x =1 là nghiệm

<=>\(\left(\sqrt{3+2x-x^2}+2\right)\left[x\left(x+1\right)-2\right]=-2\left(x+1\right)\left(x-1\right)\)

\(\left[{}\begin{matrix}-1\le x< 1\Rightarrow VP< 0;VT\ge0\rightarrow Vonghiem\\1< x\le3\Rightarrow VP>0;VT< 0\rightarrow Vonghiem\end{matrix}\right.\)

x =1 là duy nhất

Answer 2

x=1