\(VT=2\left(x^2-2.x.\frac{11}{4}+\frac{121}{16}\right)+\frac{47}{8}>0\)
=> \(VP>0\)=> x>1
pt <=> \(2\left(x^2-6x+9\right)=3\sqrt[3]{4x-4}-\left(x+3\right)\)
<=> \(2\left(x-3\right)^2=\frac{27\left(4x-4\right)-\left(x+3\right)^3}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\)
<=> \(2\left(x-3\right)^2=\frac{-\left(x+15\right)\left(x-3\right)^2}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\)
<=> \(\left(x-3\right)^2\left(2+\frac{x+15}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\right)=0\)
x>1 => $\(2+\frac{x+15}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}>0\)
pT <=> \(\left(x-3\right)^2=0\)
<=> x=3
Dễ dàng chứng minh được: \(x\ge1\)
Ta có:
\(2x^2-11x+21=2\left(x-1\right)^2+8-7x+11\)
\(\ge8\left(x-1\right)-7x+11=\left(x-1\right)+2+2\ge3\sqrt[3]{4\left(x-1\right)}\)
Dấu = xảy ra khi: \(x=3\)
