a: ĐKXĐ: x>=1
Ta có: \(\frac23\cdot\sqrt{9x-9}+\frac13\cdot\sqrt{4x-4}-\sqrt{16x-16}=\frac{10}{3}\)
=>\(\frac23\cdot3\sqrt{x-1}+\frac13\cdot2\sqrt{x-1}-4\cdot\sqrt{x-1}=\frac{10}{3}\)
=>\(2\sqrt{x-1}+\frac23\sqrt{x-1}-4\sqrt{x-1}=\frac{10}{3}\)
=>\(-\frac43\cdot\sqrt{x-1}=\frac{10}{3}\)
=>\(\sqrt{x-1}=\frac{10}{3}:\frac{-4}{3}=-\frac{10}{4}=-\frac52\) (vô lý)
=>x∈∅
b: ĐKXĐ: x∈R
\(\sqrt{9x^2-30x+25}=2\)
=>\(\sqrt{\left(3x-5\right)^2}=2\)
=>|3x-5|=2
=>\(\left[\begin{array}{l}3x-5=2\\ 3x-5=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=7\\ 3x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac73\\ x=1\end{array}\right.\)
d: ĐKXĐ: x>=0
\(2x-\sqrt{x}-15=0\)
=>\(2x-6\sqrt{x}+5\sqrt{x}-15=0\)
=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}+5\right)=0\)
=>\(\sqrt{x}-3=0\)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
