Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ

\(\sqrt{12-\frac{12}{x^2}}+\sqrt{x^2-\frac{12}{x^2}}=x^2\)

\(pt\Leftrightarrow\sqrt{12-\frac{12}{x^2}}-3+\sqrt{x^2-\frac{12}{x^2}}-1=x^2-4\)

\(\Leftrightarrow\frac{12-\frac{12}{x^2}-9}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{x^2-\frac{12}{x^2}-1}{\sqrt{x^2-\frac{12}{x^2}}+1}=x^2-4\)

\(\Leftrightarrow\frac{\frac{3x^2-12}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^4-x^2-12}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x^2-4\right)=0\)

\(\Leftrightarrow\frac{\frac{3\left(x-2\right)\left(x+2\right)}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{\left(x-2\right)\left(x+2\right)\left(x^2+3\right)}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(\frac{\frac{3}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^2+3}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-1\right)=0\)

SUy ra x=±2

\(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)+\(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)=12

Đặt  \(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)=a                                   (a>0)

=> \(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)= \(a^2\)

^{Chị QA 114 đấy}

ĐKXĐ: ...

\(\sqrt{x^2-x-30}-3\sqrt{x+5}-2\sqrt{x-6}=-6\)

\(\Leftrightarrow\sqrt{\left(x+5\right)\left(x-6\right)}-3\sqrt{x+5}-2\sqrt{x-6}=-6\)(*)

đặt \(\sqrt{x+5}=a\ge0;\sqrt{x-6}=b\ge0\)

\(\text{pt(*)}\Leftrightarrow ab-3a-2b=-6\\ \Leftrightarrow\Leftrightarrow ab-3a-2b+6=0\\ \Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\\ \Leftrightarrow\left(a-2\right)\left(b-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=2\\\sqrt{x-6}=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+5=4\\x-6=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=15\left(tm\right)\end{matrix}\right.\)

cái nằm dưới căn pt đc (7x-4)(x^2-x+3) , (7x-4)+(x^2-x+3)=x^2+6x-1 ,đặt ẩn phụ mà triển

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