ĐK: \(x\ge\frac{1}{5}\)
\(PT\Leftrightarrow\left[x+1-\sqrt{5x-1}\right]+\left[x+1-\sqrt[3]{9-x}\right]+2x^2+x-3=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{x+1+\sqrt{5x-1}}+\frac{\left(x-1\right)\left(x^2+4x+8\right)}{\left(x+1\right)^2+\left(x+1\right)\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}+\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\frac{x-2}{x+1+\sqrt{5x-1}}+....\right]=0\)
=> x=1
Ta chứng minh vế trong ngoặc >0
Từ ĐK ta có \(2x+3+\frac{x-2}{x+1+\sqrt{5x-1}}>\frac{17}{5}+\left(\frac{1}{5}-2\right)=\frac{8}{5}>0\)
\(ĐK:x\ge\frac{1}{5}\)
\(\sqrt{5x-1}+\sqrt[3]{9-x}=2x^2+3x-1\)
\(\Leftrightarrow\left(\sqrt{5x-1}-2\right)+\left(\sqrt[3]{9-x}-2\right)=2x^2+3x-5\)
\(\Leftrightarrow\frac{5\left(x-1\right)}{\sqrt{5x-1}+2}-\frac{x-1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}=\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5+\frac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}-\frac{5}{\sqrt{5x-1}+2}\right)=0\)
Với điều kiện \(x\ge\frac{1}{5}\)thì \(2x+5-\frac{5}{\sqrt{5x-1}+2}\ge2.\frac{1}{5}+5-\frac{5}{0+2}=\frac{29}{10}>0\)
Suy ra \(2x+5+\frac{1}{\sqrt[3]{\left(9-x\right)^2}+2\sqrt[3]{9-x}+4}-\frac{5}{\sqrt{5x-1}+2}>0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy phương trình có một nghiệm duy nhất là x = 1
