a, đk : x >= 1
\(\sqrt{5x-1}-\sqrt{3x-2}-\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{5x-1}-3-\sqrt{3x-2}+2-\sqrt{x-1}+1=0\)
\(\Leftrightarrow\dfrac{5x-1-9}{\sqrt{5x-1}+3}-\dfrac{3x-2-4}{\sqrt{3x-2}+2}-\dfrac{x-1-1}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\dfrac{5x-10}{\sqrt{5x-1}+3}-\dfrac{3x-6}{\sqrt{3x-2}+2}-\dfrac{x-2}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{5}{\sqrt{5x-1}+3}-\dfrac{3}{\sqrt{3x-2}+2}-\dfrac{1}{\sqrt{x-1}+1}\ne0\right)=0\Leftrightarrow x=2\)
b, tương tự
c.
ĐKXĐ: \(x\ge1\)
\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1=4x-3+2\sqrt{3x^2-5x+2}\)
\(\Leftrightarrow x+2=2\sqrt{3x^2-5x+2}\)
\(\Leftrightarrow x^2+4x+4=4\left(3x^2-5x+2\right)\)
\(\Leftrightarrow11x^2-24x+4=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{11}< 1\left(loại\right)\end{matrix}\right.\)
d.
ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
\(\sqrt{3x+4}=\sqrt{2x+1}+\sqrt{x+3}\)
\(\Leftrightarrow3x+4=3x+4+2\sqrt{2x^2+7x+3}\)
\(\Leftrightarrow2\sqrt{2x^2+7x+3}=0\)
\(\Leftrightarrow2x^2+7x+3=0\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\left(loại\right)\end{matrix}\right.\)
