a) \(\sqrt{2x-3}=3\Leftrightarrow\left(\sqrt{2x-3}\right)^2=3^2\Leftrightarrow2x-3=9\Leftrightarrow2x=3+9=12\Leftrightarrow x=12:2=6\)
a) \(\Leftrightarrow2x-3=9\Leftrightarrow x=6\)
b)\(\Leftrightarrow x^2-4x+13=9\) \(\Leftrightarrow x^2-4x+4=0\) \(\Leftrightarrow\left(x-2\right)^2=0\) \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c) \(\Leftrightarrow2=4\sqrt{x-3}\) \(\Leftrightarrow4=16x-48\) \(\Leftrightarrow16x=52\Leftrightarrow x=3,25\)
d)\(\Leftrightarrow x^2-3x+2=x-1\) \(\Leftrightarrow x^2-4x+3=0\) \(\Leftrightarrow x^2-4x+4-1=0\)
\(\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow x-2=1\Leftrightarrow x=3\)
e) \(\Leftrightarrow2x+3=27\Leftrightarrow2x=24\Leftrightarrow x=12\)
f) \(\dfrac{1}{5}\sqrt{25.\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9.\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(\Leftrightarrow\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(\Leftrightarrow9-\sqrt{x+2}=0\) \(\Leftrightarrow9=\sqrt{x+2}\Leftrightarrow81=x+2\Leftrightarrow x=79\)
g) \(\Leftrightarrow x^2-4-\sqrt{x^2-4}=0\)
\(\Leftrightarrow\sqrt{x^2-4}.\left(\sqrt{x^2-4}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2-4}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-4=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm\sqrt{5}\end{matrix}\right.\)
h) đkxđ \(x\ge-\sqrt{5}\)
\(\Leftrightarrow2\sqrt{x^2+5}-\dfrac{1}{2}\sqrt{4.\left(x^2+5\right)}+\dfrac{1}{3}\sqrt{9.\left(x^2+5\right)}=6\)
\(\Leftrightarrow2\sqrt{x^2+5}-\sqrt{x^2+5}+\sqrt{x^2+5}=6\)
\(\Leftrightarrow2\sqrt{x^2+5}=6\)
\(\Leftrightarrow\sqrt{x^2+5}=3\Leftrightarrow x^2+5=9\) \(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
Vậy...
a: Ta có: \(\sqrt{2x-3}=3\)
\(\Leftrightarrow2x-3=9\)
\(\Leftrightarrow2x=12\)
hay x=6
b: Ta có: \(\sqrt{x^2-4x+13}=3\)
\(\Leftrightarrow x^2-4x+13=9\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
c: Ta có: \(\dfrac{2}{\sqrt{x-3}}=4\)
\(\Leftrightarrow\sqrt{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x-3=\dfrac{1}{4}\)
hay \(x=\dfrac{13}{4}\)
d: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
\(\Leftrightarrow x^2-3x+2=x-1\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\sqrt[3]{2x+3}=3\)
\(\Leftrightarrow2x+3=27\)
\(\Leftrightarrow2x=24\)
hay x=12
f: Ta có: \(\dfrac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\Leftrightarrow\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}=-9\)
\(\Leftrightarrow\sqrt{x+2}=9\)
\(\Leftrightarrow x+2=81\)
hay x=79
g: Ta có: \(\sqrt{x^2-4}=x^2-4\)
\(\Leftrightarrow x^2-4=\left(x^2-4\right)^2\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
h: Ta có: \(2\sqrt{x^2+5}-\dfrac{1}{2}\sqrt{4x^2+20}+\dfrac{1}{3}\sqrt{9x^2+45}=6\)
\(\Leftrightarrow2\sqrt{x^2+5}-\sqrt{x^2+5}+\sqrt{x^2+5}=6\)
\(\Leftrightarrow x^2+5=9\)
hay \(x\in\left\{2;-2\right\}\)
