Question

[Bành Thụy Hóii]

Giải phương trình

a, \(\sqrt{x-1+4\sqrt{x-5}}+\sqrt{11+x+8\sqrt{x-5}}=0\)

b, \(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=\sqrt{8}\)

c. \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)

d, \(\sqrt{\dfrac{x^2+x+1}{x}}+\sqrt{\dfrac{x}{x^2+x+1}}=\dfrac{7}{4}\)

Answer 1

[Hồng Phúc]

b, ĐKXĐ: \(x\ge\frac{5}{2}\)

\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)

\(\Leftrightarrow\sqrt{2x-5}=3\)

\(\Leftrightarrow x=7\left(tm\right)\)

Answer 2

[Hồng Phúc]

a, ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)

\(\Leftrightarrow2\sqrt{x-5}+6=0\)

\(\Leftrightarrow\sqrt{x-5}=-3\)

Phương trình vô nghiệm