a.
ĐKXĐ: \(x;y\ge0\)
Trừ vế cho vế:
\(x^2+2x-y^2-2y+\sqrt{x}-\sqrt{y}+\sqrt{x^2+2x+22}-\sqrt{y^2+2y+22}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+2\right)+\dfrac{x-y}{\sqrt{x}+\sqrt{y}}+\dfrac{\left(x-y\right)\left(x+y+2\right)}{\sqrt{x^2+2x+22}+\sqrt{y^2+2y+22}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+2+\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{x+y+2}{\sqrt{x^2+2x+22}+\sqrt{y^2+2y+22}}\right)=0\)
\(\Leftrightarrow x=y\)
Thay vào pt đầu:
\(\Rightarrow\sqrt{x^2+2x+22}-\sqrt{x}=x^2+2x+1\)
\(\Leftrightarrow x^2-1+\sqrt{x}-1+2x+3-\sqrt{x^2+2x+22}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\dfrac{x-1}{\sqrt{x}+1}+\dfrac{\left(x-1\right)\left(3x+13\right)}{2x+3+\sqrt{x^2+2x+22}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1+\dfrac{1}{\sqrt{x}+1}+\dfrac{3x+13}{2x+3+\sqrt{x^2+2x+22}}\right)=0\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
b. Bạn kiểm tra lại đề bài
b.
Từ \(xy-x-y=1\Rightarrow x+1=y\left(x-1\right)\) (1)
\(4x^3-12x^2+9x=-y^3+6y+3+4\)
\(\Leftrightarrow4x^3-12x^2+9x=-y^3+6y+3\left(xy-x-y\right)+4\)
\(\Leftrightarrow4x^3-12x^2+12x-4=-y^3+3y\left(x+1\right)\) (2)
\(\Leftrightarrow4\left(x-1\right)^3=-y^3+3y^2\left(x-1\right)\) (thế (1) vào vế phải của (2))
Đặt \(x-1=z\)
\(\Rightarrow4z^3-3zy^2+y^3=0\)
\(\Leftrightarrow\left(2z-y\right)^2\left(z+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2z\\y=-z\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=2x-2\\y=1-x\end{matrix}\right.\)
Thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}x\left(2x-2\right)-x-\left(2x-2\right)=1\\x\left(1-x\right)-x-\left(1-x\right)=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
