Question

[Clgt]

\(\sqrt{30-\frac{5}{x^2}}+\sqrt{6x^2-\frac{5}{x^2}}=6x^2\)

Answer 1

[Phạm Minh Quang]

\(\sqrt{30-\frac{5}{x^2}}+\sqrt{6x^2-\frac{5}{x^2}}=6x^2\)ĐKXĐ:\(\left\{{}\begin{matrix}30-\frac{5}{x^2}\ge0\\6x^2-\frac{5}{x^2}\ge0\\x\ne0\end{matrix}\right.\)(*)

PT\(\Leftrightarrow\sqrt{30-\frac{5}{x^2}}-5+\sqrt{6x^2-\frac{5}{x^2}}-1=6x^2-6\)

\(\Leftrightarrow\frac{5-\frac{5}{x^2}}{\sqrt{30-\frac{5}{x^2}}+5}+\frac{6x^2-6-\frac{5}{x^2}+5}{\sqrt{6x^2-\frac{5}{x^2}}+1}=6\left(x^2-1\right)\)

\(\Leftrightarrow\frac{5\left(x^2-1\right)}{x^2\sqrt{.....}}+\frac{\left(x^2-1\right)\left(6+\frac{5}{x^2}\right)}{\sqrt{....}}-6\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{5}{x^2\sqrt{...}}+\frac{6+\frac{5}{x^2}}{\sqrt{...}}-6\right)=0\)

gấp gáp quá thì xài tạm cách này đi vế sau thử chứng minh vô nghiệm nhé

Answer 2

[Nguyễn Việt Lâm]

\(\Leftrightarrow\sqrt{30-\frac{30}{6x^2}}+\sqrt{6x^2-\frac{30}{6x^2}}=6x^2\)

Đặt \(6x^2=a>0\)

\(\sqrt{30-\frac{30}{a}}+\sqrt{a-\frac{30}{a}}=a\)

\(\sqrt{a-\frac{30}{a}}=t\Rightarrow\left\{{}\begin{matrix}\frac{30}{a}=a-t^2\\30=a^2-at^2\end{matrix}\right.\)

\(\sqrt{a^2-at^2-a+t^2}+t=a\)

\(\Leftrightarrow\sqrt{a^2-at^2-a+t^2}=a-t\) (\(a\ge t\))

\(\Rightarrow a^2-at^2-a+t^2=a^2-2at+t^2\)

\(\Leftrightarrow at^2-2at-a=0\)

\(\Leftrightarrow a\left(t-1\right)^2=0\Rightarrow t=1\)

\(\Rightarrow a^2-a-30=0\)

\(\Rightarrow\left[{}\begin{matrix}a=6\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow6x^2=6\Rightarrow x=\pm1\)