Question

Answer 1

1: ĐKXĐ: 12-x>=0

=>x<=12

\(\sqrt[3]{x+24}+\sqrt{12-x}=6\)

=>\(\sqrt[3]{x+24}-3+\sqrt{12-x}-3=6-6=0\)

=>\(\frac{x+24-27}{\sqrt[3]{\left(x+24\right)^2}+3\cdot\sqrt[3]{x+24}+9}+\frac{12-x-9}{\sqrt{12-x}+3}=0\)

=>\(\frac{x-3}{\sqrt[3]{\left(x+24\right)^2}+3\cdot\sqrt[3]{x+24}+9}+\frac{3-x}{\sqrt{12-x}+3}=0\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x+24\right)^2}+3\cdot\sqrt[3]{x+24}+9}+\frac{-1}{\sqrt{12-x}+3}=0\right)\)

=>x-3=0

=>x=3(nhận)