Bài 2:
a: ĐKXĐ: x>=-9
\(\sqrt{x+9}=3\)
=>\(x+9=3^2=9\)
=>x=9-9=0
b: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2-4x+1}=3\)
=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot1+1^2}=3\)
=>\(\sqrt{\left(2x-1\right)^2}=3\)
=>|2x-1|=3
=>\(\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2+4x+4}=4\)
=>\(\sqrt{x^2+2\cdot x\cdot2+2^2}=4\)
=>\(\sqrt{\left(x+2\right)^2}=4\)
=>|x+2|=4
=>\(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
d: ĐKXĐ: x>=0
\(5\sqrt{12x}-4\sqrt{3x}=14-2\sqrt{48x}\)
=>\(5\cdot2\sqrt{3x}-4\sqrt{3x}=14-2\cdot4\sqrt{3x}\)
=>\(6\sqrt{3x}+8\sqrt{3x}=14\)
=>\(14\sqrt{3x}=14\)
=>\(\sqrt{3x}=1\)
=>3x=1
=>\(x=\dfrac{1}{3}\left(nhận\right)\)
e: ĐKXĐ: x>=-5
\(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\)
=>\(2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
=>\(-\sqrt{x+5}+4\sqrt{x+5}=6\)
=>\(3\sqrt{x+5}=6\)
=>\(\sqrt{x+5}=2\)
=>x+5=4
=>x=-1(nhận)
