ĐK: \(x\ge-1\).

\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)

\(\Leftrightarrow\sqrt[3]{x-2}-1+\sqrt{x+1}-2=0\)

\(\Leftrightarrow\frac{x-2-1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x+1-4}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2+\sqrt[3]{x-2}+1}}+\frac{1}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)(thỏa mãn)

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Giải phương trình sau: ∛(x-2) + √(x+1)=3

A = \(\left(\frac{1}{\sqrt{x}-3}+\frac{1}{\sqrt{x}+3}\right).\left(1-\frac{3}{\sqrt{x}}\right)\)

\(A=\left(\frac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\left(\frac{\sqrt{x}-3}{\sqrt{x}}\right)\)

\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}.\frac{1}{\sqrt{x}}\)

\(A=\frac{2}{\sqrt{x}+3}.1=\frac{2}{\sqrt{x}+3}\)

\(B=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}+\frac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}-2\right)\) \(ĐKXĐ:x\ne4;x\ge0\)

\(B=\left(\frac{\sqrt{x}.\left(\sqrt{x}+2\right)+1.\left(\sqrt{x}-2\right)+6-7\sqrt{x}}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+2\right)\)

\(B=\frac{x+2\sqrt{x}+\sqrt{x}-2+6-7\sqrt{x}}{\sqrt{x}-2}\)

\(B=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}\)

\(B=\frac{\left(\sqrt{x}\right)^2-2.2.\sqrt{x}+2^2}{\sqrt{x}-2}\)

\(B=\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}\)

\(B=\sqrt{x}-2\)

Vậy...

mình ghi sai đề \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(\frac{\left(3\sqrt{2}-\sqrt{10}\right)\sqrt{6+2\sqrt{5}}+\left(3\sqrt{2}+\sqrt{10}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(\frac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)+\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)

\(\frac{3\sqrt{10}-5\sqrt{2}+3\sqrt{2}-\sqrt{10}+3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}}{\sqrt{2}}\)

\(\frac{4\sqrt{10}}{\sqrt{2}}=4\sqrt{5}\)

gi day