Giải phương trình a. (x-2)^2+(x+3).(x+3)=6 b)(x+3)^2+(4+x).(4+x)=10

Đáp án:

$\begin{array}{l}
a){\left( {x – 2} \right)^2} + \left( {x + 3} \right)\left( {x + 3} \right) = 6\\
 \Rightarrow {x^2} – 4x + 4 + {x^2} + 6x + 9 = 6\\
 \Rightarrow 2{x^2} + 2x + 7 = 0\\
 \Rightarrow {x^2} + x + \dfrac{7}{2} = 0\\
 \Rightarrow {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{13}}{4} = 0\\
 \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{{13}}{4} = 0\left( {\text{vô}\,\text{nghiệm}} \right)\\
\text{Vậy}\,pt\,\text{vô}\,\text{nghiệm}\\
b){\left( {x + 3} \right)^2} + \left( {4 + x} \right)\left( {4 + x} \right) = 10\\
 \Rightarrow {x^2} + 6x + 9 + {x^2} + 8x + 16 = 10\\
 \Rightarrow 2{x^2} + 14x + 15 = 0\\
 \Rightarrow {x^2} + 7x + \dfrac{{15}}{2} = 0\\
 \Rightarrow {x^2} + 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} – \dfrac{{19}}{4} = 0\\
 \Rightarrow {\left( {x + \dfrac{7}{2}} \right)^2} = \dfrac{{19}}{4}\\
 \Rightarrow {\left( {x + \dfrac{7}{2}} \right)^2} = {\left( {\dfrac{{\sqrt {19} }}{2}} \right)^2}\\
 \Rightarrow x = \dfrac{{ – 7 \pm \sqrt {19} }}{2}
\end{array}$