Đặt \(\hept{\begin{cases}\sqrt{x-4}=a\\\sqrt{x+4}=b\end{cases}}\)

=> \(\hept{\begin{cases}a^2+b^2=2x\\b^2-a^2=8\\ab=\sqrt{x^2-16}\end{cases}}\)

Từ đó thì PT ban đầu thành

a + b = 2ab + a² + b² - 12

<=> (a + b)² - (a + b) - 12 = 0

<=> \(\hept{\begin{cases}\left(a+b\right)=4\\\left(a+b\right)=-3\left(loai\right)\end{cases}}\)

Tới đây thì đơn giản rồi bạn làm tiếp nhé

\(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)

\(\Leftrightarrow\left(2\sqrt{2x+4}+4\sqrt{2-x}\right)^2=\left(\sqrt{9x^2+16}\right)^2\)

\(\Leftrightarrow4\left(2x+4\right)+16\left(2-x\right)+16\sqrt{2x+4}\sqrt{2-x}=9x^2+16\)

\(\Leftrightarrow4.2\left(4-x^2\right)+16\sqrt{2\left(4-x^2\right)}=x^2+8x\)

Đặt \(\sqrt{2\left(4-x^2\right)}=a\)

\(\Rightarrow4a^2+16a=x^2+8x\)

\(\Leftrightarrow\left(2a-x\right)\left(2a+x+8\right)=0\)

Làm nốt

\(\sqrt{2x^2+16x+18}+\sqrt{x^2+1}=2x+4\left(1\right)\)

\(ĐK:x\in R\)

\(pt\left(1\right)\Leftrightarrow2x^2+16x+18+x^2+1+2\sqrt[]{(2x^2+16x+18)\left(x^2+1\right)}=4x^2+16x+16\)

\(\Leftrightarrow3+2\sqrt{(2x^2+16x+18)\left(x^2+1\right)}=x^2\)

\(\Leftrightarrow2\sqrt{(2x^2+16x+8)\left(x^2+1\right)}=x^2-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3\ge0\\4\left(2x^2+16x+8\right)\left(x^2+1\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{3}\le x\le\sqrt{3}\\4\left(2x^4+16x^3+10x^2+16x+8\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow7x^4+64x^3+46x^2+64x+23=0\)

Đặt \(a=\sqrt{x+4}+\sqrt{x-4}\left(a>0\right)\)

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

\(\Leftrightarrow a^2=2x+2\sqrt{x^2-16}\)

\(\Leftrightarrow a^2-12=2x-12+2\sqrt{x^2-16}\)

Do đó \(pt\Leftrightarrow a=a^2-12\)

\(\Leftrightarrow a^2-a-12=0\)

\(\Leftrightarrow\left(a-4\right)\left(a+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+\sqrt{x-4}=4\\\sqrt{x+4}+\sqrt{x-4}=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\varnothing\end{matrix}\right.\)

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