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:

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