a. ĐKXĐ: $5\geq x\geq \frac{3}{2}$
PT $\Leftrightarrow 5-x=2x-3$
$\Leftrightarrow x=\frac{8}{3}$ (thỏa mãn)
Vậy $x=\frac{8}{3}$
b. ĐKXĐ: $x\geq \frac{-3}{8}$
PT $\Leftrightarrow 8x+3=19^2$
$\Leftrightarrow x=\frac{179}{4}$
c. ĐKXĐ: $x\geq \frac{-9}{5}$
PT \(\left\{\begin{matrix} x+1\geq 0\\ 5x+9=(x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -1\\ x^2-3x-8=0\end{matrix}\right.\Rightarrow x=\frac{3+\sqrt{41}}{2}\)
d. ĐKXĐ: $x\geq -1$
PT \(\Leftrightarrow \sqrt{9}.\sqrt{x+1}-2\sqrt{x+1}+\sqrt{36}.\sqrt{x+1}=49\)
\(\Leftrightarrow 3\sqrt{x+1}-2\sqrt{x+1}+6\sqrt{x+1}=49\)
\(\Leftrightarrow 7\sqrt{x+1}=49\Leftrightarrow \sqrt{x+1}=7\)
\(\Leftrightarrow x+1=49\Leftrightarrow x=48\)
e. ĐKXĐ: $x\geq \frac{3}{2}$
PT \(\Leftrightarrow \sqrt{9}.\sqrt{2x-3}-5\sqrt{4}.\sqrt{2x-3}+2\sqrt{16}.\sqrt{2x-3}=49\)
\(\Leftrightarrow 3\sqrt{2x-3}-10\sqrt{2x-3}+8\sqrt{2x-3}=49\)
\(\Leftrightarrow \sqrt{2x-3}=49\Leftrightarrow 2x-3=2401\Leftrightarrow x=1202\)
f.
PT $\Leftrightarrow \sqrt{(x-4)^2}=10$
$\Leftrightarrow |x-4|=10$
$\Leftrightarrow x-4=\pm 10$
$\Leftrightarrow x=14$ hoặc $x=-6$
f: Ta có: \(\sqrt{x^2-8x+16}=10\)
\(\Leftrightarrow\left|x-4\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=10\\x-4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=14\\x=-6\end{matrix}\right.\)
