a) ĐKXĐ: \(x\ge-1\)
Ta có: \(\sqrt{\left(\sqrt{x+1}\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x+1}\right|=2\)
\(\Leftrightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x+1=4\)
hay x=3(nhận)
Vậy: x=3
b)ĐKXĐ: \(x\ge\frac{2}{3}\)
Ta có: \(\sqrt{3x-2}-5=3\)
\(\Leftrightarrow\sqrt{3x-2}=3+5=8\)
\(\Leftrightarrow3x-2=64\)
\(\Leftrightarrow3x=66\)
hay x=22(nhận)
Vậy: x=22
c)ĐKXĐ: \(x\in R\)
Ta có: \(\sqrt{9x^2-6x+1}=2+\sqrt{3}\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}=2+\sqrt{3}\)
\(\Leftrightarrow\left|3x-1\right|=2+\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2+\sqrt{3}\\1-3x=2+\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-1-2-\sqrt{3}=0\\1-3x-2-\sqrt{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3-\sqrt{3}=0\\-3x-1-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3+\sqrt{3}\\-3x=1+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{3}}{3}\left(nhận\right)\\x=\frac{-1-\sqrt{3}}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3+\sqrt{3}}{3};\frac{-1-\sqrt{3}}{3}\right\}\)
a) \(\sqrt{\left(\sqrt{x+1}\right)^2}\)=2, x>-1
⇔\(\left|\sqrt{x+1}\right|\)=2
⇔x+1 =4
⇔x=3. (tmđk)
Vậy phương trình có nghiệm là S= \(\left\{3\right\}\).
b) \(\sqrt{3x-2}\)-5=3 , x≥ \(\frac{2}{3}\)
⇔\(\sqrt{3x-2}\) =8
⇔3x-2 =64
⇔x=22.(tmđk)
Vây phương trình có nghiệm là S=\(\left\{22\right\}\).
c) \(\sqrt{9x^2-6x+1}\)= 2+\(\sqrt{3}\)
⇔\(\sqrt{\left(3x-1\right)^2}\) =2+\(\sqrt{3}\)
⇔\(\left|3x-1\right|\)= 2+\(\sqrt{3}\)
⇔\(\left[{}\begin{matrix}3x-1=2+\sqrt{3}\\3x-1=-2-\sqrt{3}\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=\frac{3+\sqrt{3}}{3}\\x=\frac{-1-\sqrt{3}}{3}\end{matrix}\right.\)
vậy nghiệm của phương trình là S=\(\left\{\frac{3+\sqrt{3}}{3};\frac{-1-\sqrt{3}}{3}\right\}\)
