Giải phương trình: a) x^2+x+5-3√x^2+x+3=0 b) x^2-2x+5-4√x^2-2x+2=0 Giúp mình nhé!!!!

Đáp án:

$\begin{array}{l}
a){x^2} + x + 5 - 3\sqrt {{x^2} + x + 3}  = 0\\
\left( {dkxd:{x^2} + x + 3 \ge 0\,luon\,dung} \right)\\
 \Rightarrow {x^2} + x + 3 - 3\sqrt {{x^2} + x + 3}  + 2 = 0\\
 \Rightarrow {\left( {\sqrt {{x^2} + x + 3} } \right)^2} - 3\sqrt {{x^2} + x + 3}  + 2 = 0\\
 \Rightarrow {\left( {\sqrt {{x^2} + x + 3} } \right)^2} - \sqrt {{x^2} + x + 3}  - 2\sqrt {{x^2} + x + 3}  + 2 = 0\\
 \Rightarrow \left( {\sqrt {{x^2} + x + 3}  - 1} \right)\left( {\sqrt {{x^2} + x + 3}  - 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt {{x^2} + x + 3}  = 1\\
\sqrt {{x^2} + x + 3}  = 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{x^2} + x + 3 = 1\\
{x^2} + x + 3 = 4
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} + x + 2 = 0\\
{x^2} + x - 1 = 0
\end{array} \right.\\
 \Rightarrow x = \frac{{ - 1 \pm \sqrt 5 }}{2}\\
b)\\
{x^2} - 2x + 5 - 4\sqrt {{x^2} - 2x + 2}  = 0\left( {txd:R} \right)\\
 \Rightarrow {x^2} - 2x + 2 - 4\sqrt {{x^2} - 2x + 2}  + 3 = 0\\
 \Rightarrow {\left( {\sqrt {{x^2} - 2x + 2} } \right)^2} - 4\sqrt {{x^2} - 2x + 2}  + 3 = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt {{x^2} - 2x + 2}  = 1\\
\sqrt {{x^2} - 2x + 2}  = 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{x^2} - 2x + 2 = 1\\
{x^2} - 2x + 2 = 9
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} - 2x + 1 = 0\\
{x^2} - 2x - 7 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 1 \pm 2\sqrt 2 
\end{array} \right.
\end{array}$