\(x^2-y^2+2x-4y-10=0\)

\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left(x-y+1-2\right)\left(x+y+1+2\right)=7\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)

Xét bảng tìm x; y là xong

a) x² + y² +2x - 4y + 5 = 0

( x² + 2x + 1 ) + ( y² - 4y + 4 ) = 0

( x + 1 )² + ( y - 2 )² = 0

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

b) \(x^2+4y^2-x-4y+\dfrac{5}{4}=0\)

\(x^2-x+\dfrac{1}{4}+4y^2-4y+1=0\)

\(\left(x-\dfrac{1}{2}\right)^2+\left(2y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\2y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Cộng vế:

\(x^3-y^3+3x^2+3y^2+4x-4y+4=0\)

\(\Leftrightarrow\left(x+1\right)^3-\left(y-1\right)^3+x-y+2=0\)

\(\Leftrightarrow\left(x-y+2\right)\left(x^2+y^2+xy+x-y+2\right)=0\)

\(\Leftrightarrow\left(x-y+2\right)\left[\left(x+\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+1\right]=0\)

\(\Leftrightarrow y=x+2\)

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x+y\ge1\\x+2y\ge2\\x+4y\ge0\end{matrix}\right.\)

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

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

Thế vào pt 2 => x;y

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+y-1}=a\ge0\\\sqrt{x+2y-2}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=x-y+1\)

Phương trình thứ nhất trở thành:

\(a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{2x+y-1}=\sqrt{x+2y-2}\Rightarrow y=x+1\)

Thay xuống pt dưới:

\(4x^2-\left(x+1\right)^2+x+4-\sqrt{3x+1}-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x^2-x+3-\sqrt{3x+1}-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x^2-3x+x+1-\sqrt{3x+1}+x+2-\sqrt{5x+4}=0\)

\(\Leftrightarrow3x\left(x-1\right)+\frac{\left(x+1\right)^2-\left(3x+1\right)}{x+1+\sqrt{3x+1}}+\frac{\left(x+2\right)^2-\left(5x+4\right)}{x+2+\sqrt{5x+4}}=0\)

\(\Leftrightarrow3x\left(x-1\right)+\frac{x\left(x-1\right)}{x+1+\sqrt{3x+1}}+\frac{x\left(x-1\right)}{x+2+\sqrt{5x+4}}=0\)

\(\Leftrightarrow x\left(x-1\right)\left(3+\frac{1}{x+1+\sqrt{3x+1}}+\frac{1}{x+2+\sqrt{5x+4}}\right)=0\)

(x -y-3)(x+y+1)=7 hết 

chi tiết đi

\(x^2-y^2+2x-4y-10=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=13\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=13\)

\(\Leftrightarrow\left(x+y+3\right)\left(x-y-1\right)=13\)

Tới đây thì đơn giản rồi nhé

pt <=> \(\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=7\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left(x+y+3\right)\left(x-y-1\right)=7\)

Mặt khác x,y>0 => x+y+3>x-y-1 và x+y+3>0

Nên ta có cặp nghiệm duy nhất sau: \(\hept{\begin{cases}x+y+3=7\\x-y-1=1\end{cases}\Leftrightarrow}\)\(\hept{\begin{cases}x+y=4\\x-y=2\end{cases}\Leftrightarrow}\)\(\hept{\begin{cases}x=3\\y=1\end{cases}}\)

Đúng rồi \(\left(x+y+3\right)\left(x-y-1\right)=7\)

Nhầm sorry nhá