a,ĐK: x≥-1

Đặt \(t=\sqrt{x^2+5x+4}\left(t\ge0\right)\)

  ⇒ \(t^2+t-6=0\)

  \(\Leftrightarrow\left(t+3\right)\left(t-2\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}t=-3\left(loại\right)\\t=2\end{matrix}\right.\)

  \(\Leftrightarrow\sqrt{x^2+5x+4}=2\)

  \(\Leftrightarrow x^2+5x+4=4\)

  \(\Leftrightarrow x\left(x+5\right)=0\)

  \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-5\left(loại\right)\end{matrix}\right.\)

b,ĐK: \(0\le x\le2\)

Ta có: \(\left(x+5\right)\left(2-x\right)=3\sqrt{x^2+3x}\)

    \(\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)     (1)

Đặt \(t=\sqrt{x^2+3x}\left(t\ge0\right)\)

  \(\Rightarrow\left(1\right)\Leftrightarrow-t^2+10-3t=0\)  

             \(\Leftrightarrow\left(t+5\right)\left(2-t\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}t=-5\left(loại\right)\\t=2\end{matrix}\right.\)

             \(\Leftrightarrow\sqrt{x^2+3x}=2\)

             \(\Leftrightarrow x^2+3x=4\)

             \(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

             \(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=1\left(tm\right)\end{matrix}\right.\)

1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)

\(t^3+2=2t-2\)

\(\Leftrightarrow t^3-2t+4=0\)

\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)

\(\Rightarrow t=-2\)

\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)

\(\Leftrightarrow x^2+5x-2=-8\)

\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)

\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)

3/ ĐKXĐ: \(\left|x\right|\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[4]{x-\sqrt{x^2-1}}=a>0\\\sqrt[4]{x+\sqrt{x^2-1}}=b>0\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}ab=1\\a+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}ab=1\\a=2-b^2\end{matrix}\right.\)

\(\Rightarrow b\left(2-b^2\right)=1\Leftrightarrow b^3-2b+1=0\)

\(\Leftrightarrow\left(b-1\right)\left(b^2+b-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}b=1\\b^2+b-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}b=1\\b=\frac{-1+\sqrt{5}}{2}\\b=\frac{-1-\sqrt{5}}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[4]{x+\sqrt{x^2-1}}=1\\\sqrt[4]{x+\sqrt{x^2-1}}=\frac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x^2-1}=1\\x+\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2-1}=\frac{7-3\sqrt{5}}{2}-x\left(vn\right)\end{matrix}\right.\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

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