a:
ĐKXĐ: x>=2
\(\Rightarrow\sqrt{4x-8}=\sqrt{x^3-8}\)
\(\Leftrightarrow x^3-8=4x-8\)
\(\Leftrightarrow x^3-4x=0\)
=>x(x-2)(x+2)=0
\(\Leftrightarrow x=2\)
b: ĐKXĐ: x>=1
\(\Leftrightarrow3\cdot2\sqrt{x-1}+9\cdot\dfrac{\sqrt{x-1}}{3}-5\sqrt{x-1}=2\)
\(\Leftrightarrow4\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\)
=>x-1=1/4
hay x=5/4(nhận)
c: \(\Leftrightarrow5x+5\sqrt{x}-2\sqrt{x}-2=5x+4\)
\(\Leftrightarrow3\sqrt{x}-2=4\)
\(\Leftrightarrow3\sqrt{x}=6\)
hay x=4
a.\(2\sqrt{x-2}=\sqrt{x^3-8}\)
\(ĐK:x\ge2\)
\(\Leftrightarrow2\sqrt{x-2}=\sqrt{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\Leftrightarrow2\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x^2+2x+4\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(2-\sqrt{x^2+2x+4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\2-\sqrt{x^2+2x+4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+2x+4=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
b.\(3\sqrt{4x-4}+9\sqrt{\dfrac{x-1}{9}}-\sqrt{25x-25}=0\)
\(\Leftrightarrow3\sqrt{4\left(x-1\right)}+9\sqrt{\dfrac{x-1}{9}}-\sqrt{25\left(x-1\right)}=0\)
\(ĐK:x\ge1\)
Đặt \(x-1=a\)
\(\Leftrightarrow3\sqrt{4a}+9\sqrt{\dfrac{a}{9}}-\sqrt{25a}=2\)
\(\Leftrightarrow6\sqrt{a}+3\sqrt{a}-5\sqrt{a}=2\)
\(\Leftrightarrow4\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=\dfrac{1}{2}\)
\(\Leftrightarrow a=\dfrac{1}{4}\)
\(\Rightarrow x-1=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{5}{4}\right\}\)
c.\(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=5x+4\) ; \(ĐK:x\ge0\)
\(\Leftrightarrow5x+5\sqrt{x}-2\sqrt{x}-2-5x-4=0\)
\(\Leftrightarrow3\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow x=4\left(tm\right)\)
Vậy \(S=\left\{4\right\}\)
