Question

[Nguyễn Hoàng]

Giải phương trình: \(2\sqrt{x^3+1}+\sqrt{\frac{x^4+16}{2}}=3x^2-6x+10\)

Answer 1

[alibaba nguyễn]

Ta có:

\(3x^2-6x+10-2\sqrt{x^3+1}-\sqrt{\frac{x^4+16}{2}}=0\)

Ta có:

\(VT=\left(x-2\right)^2+2x^2-2x+6-2\sqrt{\left(x+1\right)\left(x^2-x+1\right)}-\sqrt{\frac{x^4+16}{2}}\)

\(\ge2x^2-2x+6-\left(x+1+x^2-x+1\right)-\sqrt{\frac{x^4+16}{2}}\)

\(\ge x^2-2x+4-\sqrt{\frac{x^4+16}{2}}\)

Ta chứng minh:

\(x^2-2x+4-\sqrt{\frac{x^4+16}{2}}\ge0\)

\(\Leftrightarrow x^2-2x+4\ge\sqrt{\frac{x^4+16}{2}}\)

\(\Leftrightarrow\left(x^2-2x+4\right)^2\ge\frac{x^4+16}{2}\)

\(\Leftrightarrow\left(x-2\right)^4\ge0\)

Vậy \(VT\ge VP\)dấu = xảy ra khi \(x=2\)