ĐKXĐ: x>=1
Ta có: \(\sqrt[3]{x-2}+\sqrt{x-1}=5\)
=>\(\sqrt[3]{x-2}-2+\sqrt{x-1}-3=5-5=0\)
=>\(\frac{x-2-8}{\sqrt[3]{\left(x-2\right)^2}+2\cdot\sqrt[3]{x-2}+4}+\frac{x-1-9}{\sqrt{x-1}+3}=0\)
=>\(\left(x-10\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+2\cdot\sqrt[3]{x-2}+4}+\frac{1}{\sqrt{x-1}+3}\right)=0\)
=>x-10=0
=>x=10(nhận)
