Question

[Lê Thành An]

giải hệ \(\sqrt{2x+3}+x^2+2x=\sqrt{2y-1}+y^2-2y\) và \(\sqrt{x-2}+\sqrt{y-1}=3\)

Answer 1

[Cô nàng Thiên Bình dễ thương]

\(\hept{\begin{cases}\sqrt{2x+3}+x^2+2x=\sqrt{2y-1}+y^2-2y\left(1\right)\\\sqrt{x-2}+\sqrt{y-1}=3\left(2\right)\end{cases}}\)

\(Đkxđ:x\ge2;y\ge1\)

\(\left(1\right)\Leftrightarrow\sqrt{2x+3}-\sqrt{2y-1}=y^2-x^2-2\left(y+x\right)\)

\(\frac{2x-2y+4}{\sqrt{2x+3}+\sqrt{2y-1}}=\left(x+y\right)\left(y-x\right)-2\left(y+x\right)\)

\(\Leftrightarrow\frac{2\left(x-y+2\right)}{\sqrt{2x+3}+\sqrt{2y-1}}+\left(x+y\right)\left(x-y+2\right)=0\)

\(\Leftrightarrow\left(x-y+2\right)\left(\frac{2}{\sqrt{2x+3}+\sqrt{2y-1}}+x+y\right)=0\)

\(\Leftrightarrow x-y+2=0\)

\(\Leftrightarrow x=y-2\)

Thay vào \(\left(2\right)\) ...................................................................