Question

a) Giải \(\left\{{}\begin{matrix}\dfrac{2}{x+3}-\dfrac{5}{y-2}=1\\\dfrac{x+4}{x+3}+\dfrac{y+3}{y-2}=4\end{matrix}\right.\)

b) Giải : \(x^2+2\sqrt{3}x-6=0\)

Answer 1

b) \(x^2+2\sqrt{3}x-6=0\)

\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)

Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}