Sửa lại \(\sqrt[3]{x+6}+1=x^2-\sqrt{x-1}\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt[3]{x+6}+\sqrt{x-1}-\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(\sqrt[3]{x+6}-2\right)+\left(\sqrt{x-1}-1\right)-\left(x^2-4\right)=0\left(1\right)\)
\(\sqrt[3]{x+6}-2=\dfrac{\left(\sqrt[3]{x+6}-2\right)\left[\left(\sqrt[3]{x+6}\right)^2+2\sqrt[3]{x+6}+4\right]}{\left(\sqrt[3]{x+6}\right)^2+2\sqrt[3]{x+6}+4}=\dfrac{x-2}{\left(\sqrt[3]{x+6}\right)^2+2\sqrt[3]{x+6}+4}\)
\(\sqrt{x-1}-1=\dfrac{\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+1\right)}{\sqrt{x-1}+1}=\dfrac{x-2}{\sqrt{x-1}+1}\)
\(x^2-4=\left(x-2\right)\left(x+2\right)\)
\(\left(1\right)\Leftrightarrow\left(x-2\right)\left[\dfrac{1}{\left(\sqrt[3]{x+6}\right)^2+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt{x-1}+1}-\left(x+2\right)\right]=0\left(2\right)\)
vì \(x\ge1\Leftrightarrow\dfrac{1}{\left(\sqrt[3]{x+6}\right)^2+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt{x-1}+1}-\left(x+2\right)\le\dfrac{1}{\left(\sqrt[3]{7}\right)^2+2\sqrt[3]{7}+4}+1-3\)
\(=\dfrac{1}{\left(\sqrt[3]{7}\right)^2+2\sqrt[3]{7}+4}-2< 0\)
\(\left(2\right)\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy phương trình cho có 1 nghiệm duy nhất \(x\in\left\{2\right\}\)
