a, Cộng vế theo vế hai phương trình ta được:

\(x^2+y^2+2xy+x+y=2\)

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

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

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

TH1: \(x+y=1\)

\(pt\left(2\right)\Leftrightarrow xy+1=-1\Leftrightarrow xy=-2\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=1\\xy=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\end{matrix}\right.\)

TH2: \(x+y=-2\)

\(pt\left(2\right)\Leftrightarrow xy-2=-1\Leftrightarrow xy=1\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=-2\\xy=1\end{matrix}\right.\Leftrightarrow x=y=-1\)

b, \(\left\{{}\begin{matrix}x^3-y^3=7\left(x-y\right)\\x^2+y^2=x+y+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2+xy-7\right)=0\\x^2+y^2=x+y+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x^2+y^2+xy=7\end{matrix}\right.\\x^2+y^2=x+y+2\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2=x+y+2\end{matrix}\right.\)

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

\(\Leftrightarrow x=y=\dfrac{1\pm\sqrt{5}}{2}\)

TH2: \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2=x+y+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=7\\\left(x+y\right)^2-2xy-x-y=2\end{matrix}\right.\)

Đặt \(x+y=u;xy=v\)

Hệ trở thành: \(\left\{{}\begin{matrix}u^2-v=7\\u^2-2v-u=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2-2\left(u^2-7\right)-u=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\u^2+u-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-7\\\left[{}\begin{matrix}u=3\\u=-4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}v=2\\u=3\end{matrix}\right.\\\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}v=2\\u=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}v=9\\u=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=9\\x+y=-4\end{matrix}\right.\left(vn\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)

Cộng vế với vế:

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

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

TH1: \(x+y=1\Rightarrow y=1-x\) thay vào pt dưới:

\(x\left(1-x\right)+x+2\left(1-x\right)-1=0\)

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

TH2: \(x+y=-4\Rightarrow y=-4-x\)

\(x\left(-4-x\right)+x+2\left(-4-x\right)-1=0\)

\(\Leftrightarrow x^2+5x+9=0\) (vô nghiệm)

a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

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

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

\(\Rightarrow a^2b^2-b^4=b-a\)

\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)

\(\Leftrightarrow a=b\)

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

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

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

Ta có: x³ + y³ = x + y
   <=> (x + y)(x² - xy + y²) - (x + y) = 0
   <=> (x + y)(x² - xy + y² - 1) = 0
   <=> x + y = 0 hay x² - xy + y² - 1 = 0
* x + y = 0 => x = -y
Thế vào pt (1), ta có 2y² - y² = 1 <=> y² = 1 <=> y = 1 hay y = -1
@ y = 1 => x = -1
@ y = -1 => x = 1
* x² - xy + y² - 1 = 0 => x² - xy + y² = 1 (3)
Lấy (1) - (3) vế theo vế, ta có: 2xy = 0 <=> x = 0 hay y = 0
@ x = 0 => y² = 1 <=> y = 1 hay y = -1
@ y = 0 => x² = 1 <=> x = 1 hay x = -1
Vậy nghiệm (x; y) = (-1, 1) ; (1; -1) ; (0; 1) ; (0; -1) ; (1; 0) ; (-1; 0)