ĐKXĐ: \(x\ge\sqrt[3]{7}\)
\(\sqrt{x^4-7}-\left(x^2-1\right)+\sqrt{x^3-7}-1=0\)
\(\Leftrightarrow\dfrac{x^4-7-\left(x^2-1\right)^2}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^3-8}{\sqrt{x^3-7}+1}=0\)
\(\Leftrightarrow\dfrac{2\left(x^2-4\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{2\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^2+2x+4}{\sqrt{x^3-7}+1}\right)=0\)
Do \(x\ge\sqrt[3]{7}>1\Rightarrow x^2>1\Rightarrow x^2-1>0\)
\(\Rightarrow\dfrac{2\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^2+2x+4}{\sqrt{x^3-7}+1}>0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
