a/ $\begin{cases}\frac{x+3}{y-4}-\frac{x-1}{y+4}+\frac{16}{y^{2}-16}=0 (1) \\ 11x-3y=1 (2)\end{cases}$
Thế $y=\frac{11x-1}{3}$ vào \((1)\):
$\frac{x+3}{\frac{11x-1}{3}-4}-\frac{x-1}{\frac{11x-1}{3}+4}+\frac{16}{\left(\frac{11x-1}{3}\right)^{2}-16}=0$
$\Leftrightarrow \frac{3\left(x+3\right)}{11x-13}-\frac{3\left(x-1\right)}{11x+11}+\frac{144}{\left(11x-13\right)\left(11x+11\right)}=0$
$\Leftrightarrow \frac{\left(x+3\right)}{11x-13}-\frac{\left(x-1\right)}{11x+11}+\frac{48}{\left(11x-13\right)\left(11x+11\right)}=0$
điều kiện: $x\neq -1, x\neq \frac{13}{11}, y\neq \pm 4$
với điều kiện trên, biến đổi phương trình:
$\left(x+3\right)\left(11x+11\right)-\left(x-1\right)\left(11x-13\right)+48=0$
$\Leftrightarrow 11x^{2}+44x+33-11x^{2}+24x-13+48=0$
$\Leftrightarrow 68x+68=0\Leftrightarrow x=-1$: hệ vô nghiệm
b/ $\begin{cases}2y^{2}+xy-x^{2}=0 (1)\\ x^{2}-xy-y^{2}+3x+7y+3=0 (2)\end{cases}$
$(1)\Leftrightarrow y^{2}+xy+y^{2}-x^{2}=0\Leftrightarrow y\left(y+x\right)+\left(y+x\right)\left(y-x\right)=0$ $\Leftrightarrow \left(y+x\right)\left(2y-x\right)=0$
$\Leftrightarrow \left[ \begin{array}{l}y = -x\\y=\frac{x}{2}\end{array} \right.$
Thế $y=-x$ vào $(2)$:
$x^{2}-x\left(-x\right)-\left(-x\right)^{2}+3x+7\left(-x\right)+3=0\Leftrightarrow x^{2}-4x+3=0$ $\Leftrightarrow \left[ \begin{array}{l}x = 1\Rightarrow y=-1\\x=3 \Rightarrow y=-3\end{array} \right.$
thế $y=\frac{x}{2}$ vào $(2)$:
$x^{2}-x\left(\frac{x}{2}\right)-\left(\frac{x}{2}\right)^{2}+3x+7\left(\frac{x}{2}\right)+3=0\Leftrightarrow \frac{x^{2}}{4}+\frac{13x}{2}+3=0$
$ \Leftrightarrow \left[ \begin{array}{l}x = -13-\sqrt{157} \Rightarrow \frac{ -13-\sqrt{157}}{2}\\x= -13+\sqrt{157} \Rightarrow y=\frac{ -13+\sqrt{157}}{2}\end{array} \right.$
c/$\begin{cases}xy+3y^{2}-x+4y-7=0 (1)\\ 2xy+y^{2}-2x-2y+1=0 (2)\end{cases}$
$(2)\Leftrightarrow y^{2}+2\left(x-1\right)y+1-2x=0\Leftrightarrow \left[ \begin{array}{l}y=1\\y = 1-2x\end{array} \right.$
Thế $y=1$ vào $(1): x+3-x+4-7=0$ Đúng $\forall x \in R$
$y=1-2x\Rightarrow (1): x\left(1-2x\right)+3\left(1-2x\right)^{2}-x+4\left(1-2x\right)-7=0$
$\Leftrightarrow x-2x^{2}+3-12x+12x^{2}-x+4-8x-7=0\Leftrightarrow 10x^{2}-20x=0$
$\Leftrightarrow x=0, x=2\Rightarrow y=1, y=-3$
d/$\begin{cases}xy-3x-2y=16 \\ x^{2}+y^{2}-2x-4y=33 \end{cases}\Leftrightarrow \begin{cases}2xy-6x-6y=32(1) \\ x^{2}+y^{2}-2x-4y=33(2) \end{cases}$
Cộng từng vế: $\left(x+y\right)^{2}-8\left(x+y\right)-65=0$
$\Leftrightarrow \left[ \begin{array}{l}x+y = -5\\ x+y=13\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}y=-x-5\\y = 13-x\end{array} \right.$ thế vào $(1)$:
$\left[ \begin{array}{l}2x\left(-x-5\right) -6x-6\left(-x-5\right)-32=0\\2x \left(13-x\right)-6x-6\left(13-x\right)-32=0\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}x^{2}+5x+1=0\\ x^{2}-13x+55=0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = \frac{-5-\sqrt{21}}{2}, x=\frac{-5+\sqrt{21}}{2}\\x=\Phi\end{array} \right.$
$\Rightarrow y=\frac{5+\sqrt{21}}{2}-5=\frac{-5+\sqrt{21}}{2}, y=\frac{5-\sqrt{21}}{2}-5=\frac{-5-\sqrt{21}}{2}$
e/ $\begin{cases}x-\frac{1}{x}=y-\frac{1}{y} (1)\\ 2y=x^{3}+1 (2)\end{cases}$
điều kiện: $x\neq 0, y\neq 0$
$(1)\Leftrightarrow x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\Leftrightarrow \left(x-y\right)\left(1+\frac{1}{xy}\right)=0$
$\Leftrightarrow \left[ \begin{array}{l}x = y\\xy =-1\end{array} \right.$
Thế $y=x$ vào $(2): 2x=x^{3}+1\Leftrightarrow x^{3}-2x+1=0$
$\Leftrightarrow \left(x^{3}-1\right)-2\left(x-1\right)=0 \Leftrightarrow \left(x-1\right)\left(x^{2}+x+1-2\right)=0$
$\Leftrightarrow x=1, x=\frac{-1-\sqrt{5}}{2}, x=\frac{-1+\sqrt{5}}{2}$
$\Rightarrow y=1, y=\frac{-1-\sqrt{5}}{2}, y=\frac{-1+\sqrt{5}}{2}$
$xy=-1\Rightarrow y=\frac{-1}{x}$ thế vào $(2): 2\left(\frac{-1}{x}\right)=x^{3}+1$
$\Leftrightarrow \left(x^{4}-x^{2}+1\right)+\left(x^{2}+x+1\right)=0 (*)$
$x^{4}-x^{2}+1= \left(x^{2}-\frac{1}{x}\right)^{2}+\frac{3}{4} \geq \frac{3}{4}$
$ x^{2}+x+1=\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4}$
$\Rightarrow \left(x^{4}-x^{2}+1\right)+\left(x^{2}+x+1\right)>\frac{3}{2}: (*)$ vô nghiệm