Bài 2:
a) \(x-3\sqrt{x}-4=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow x+\sqrt{x}-4\sqrt{x}-4=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-4=0\\\sqrt{x}+1\ge1>0\forall x\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=4^2\)
\(\Leftrightarrow x=16\left(tm\right)\)
b) \(\sqrt{2x-1}+\sqrt{x-1}=5\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{2x-1}+\sqrt{x-1}-5=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\dfrac{2x-1-9}{\sqrt{2x-1}+3}+\dfrac{x-1-4}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\dfrac{2\left(x-5\right)}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\right)=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\left(tm\right)\)
c) \(x^2+2x+7=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\) (ĐK: \(x\ge-3\))
\(\Leftrightarrow x^2+1+2x+6=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(x^2+1\right)+2\left(x+3\right)=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{x^2+1}\\b=\sqrt{x+3}\end{matrix}\right.\)
\(\Leftrightarrow a^2+2b^2=3ab\)
\(\Leftrightarrow a^2-3ab+2b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)
TH1: a=b
\(\Leftrightarrow\sqrt{x^2+1}=\sqrt{x+3}\)
\(\Leftrightarrow x^2+1=x+3\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
TH2: \(a=2b\)
\(\Leftrightarrow\sqrt{x^2+1}=2\sqrt{x+3}\)
\(\Leftrightarrow x^2+1=4\left(x+3\right)\)
\(\Leftrightarrow x^2-4x-11=0\)
\(\Leftrightarrow x=2\pm\sqrt{5}\left(tm\right)\)
