Giải phương trình a, √ -6/4+x=5 b,√4x-20-3√x-5/5=√1-x c,√3x²-4x+3=1-2x d,√9x+9 + √4x+4=√x+1

Đáp án:

$\begin{array}{l}
a)\sqrt {\dfrac{{ - 6}}{{4 + x}}}  = 5\\
 \Leftrightarrow \dfrac{{ - 6}}{{4 + x}} = 25\\
 \Leftrightarrow 4 + x =  - \dfrac{6}{{25}}\\
 \Leftrightarrow x =  - \dfrac{{106}}{{25}}\\
Vậy\,x =  - \dfrac{{106}}{{25}}\\
b)Dkxd:\left\{ \begin{array}{l}
x \ge 5\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 5\\
x \le 1
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)\sqrt {3{x^2} - 4x + 3}  = 1 - 2x\left( {dk:x \le \dfrac{1}{2}} \right)\\
 \Leftrightarrow 3{x^2} - 4x + 3 = 1 - 4x + 4{x^2}\\
 \Leftrightarrow {x^2} = 2\\
 \Leftrightarrow x =  - \sqrt 2 \left( {do:x \le \dfrac{1}{2}} \right)\\
Vậy\,x =  - \sqrt 2 \\
d)Dkxd:x \ge  - 1\\
\sqrt {9x + 9}  + \sqrt {4x + 4}  = \sqrt {x + 1} \\
 \Leftrightarrow 3\sqrt {x + 1}  + 2\sqrt {x + 1}  - \sqrt {x + 1}  = 0\\
 \Leftrightarrow 4\sqrt {x + 1}  = 0\\
 \Leftrightarrow x =  - 1\left( {tmdk} \right)\\
Vậy\,x =  - 1
\end{array}$