Question

Giải hệ phương trình:

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

và \(x^3-y^3+5=xy+\sqrt{x-1}\)

Answer 1

ĐKXĐ: \(x\ge1;y\ge\frac{1}{2}\)

\(x^3-y^3+\sqrt{x+y-1}-\sqrt{2y-1}=0\)

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

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

\(\Leftrightarrow x=y\)

Thay xuống dưới:

\(5=x^2+\sqrt{x-1}\)

\(\Leftrightarrow x^2-4+\sqrt{x-1}-1=0\)

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

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

\(\Rightarrow x=2\Rightarrow y=2\)