Giải hệ phương trình

e, $\left\{\begin{array}{l}\frac{4}{x+y}+\frac{1}{y-1}=5\\ \frac{1}{x+y}-\frac{2}{y-1}=-1\end{array}\right.$

f, $\left\{\begin{array}{l}4\sqrt{x}-3\sqrt{y}=4\\ 2\sqrt{x}+\sqrt{y}=2\end{array}\right.$

 

c, Điều kiện $y\ne 0$ . Đặt $t=\frac{1}{y}$ , hệ phương trình đã cho trở thành$\left\{\begin{array}{l}x+t=\frac{-1}{2}\\ 2x-3t=\frac{-7}{2}\end{array}\right.\iff \left\{\begin{array}{l}t=\frac{-1}{2}-x\\ 2x-3(\frac{-1}{2}-x)=\frac{-7}{2}\end{array}\right.\iff \left\{\begin{array}{l}t=\frac{-1}{2}-x\\ 5x=-5\end{array}\right.\iff \left\{\begin{array}{l}x=-1\\ t=\frac{1}{2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=-1\\ y=2\end{array}\right.$(thỏa mãn)

Vậy hệ có nghiệm duy nhất là $\left(x;y\right)=\left(-1;2\right)$ .

d, $\left(I\right)\left\{\begin{array}{l}\frac{3x}{x-1}-\frac{2}{y+2}=4\\ \frac{2x}{x-1}+\frac{1}{y+2}=5\end{array}\right.$ ĐK $x\ne 1;y\ne -2$

Đặt $\left\{\begin{array}{l}\frac{x}{x-1}=a\\ \frac{1}{y+2}=b\end{array}\right.$Khi đó hệ phương trình (I) trở thành:

$\left\{\begin{array}{l}3a-2b=4\\ 2a+b=5\end{array}\right.\iff \left\{\begin{array}{l}3a-2b=4\\ 4a+2b=10\end{array}\right.\iff \left\{\begin{array}{l}7a=14\\ 2a+b=5\end{array}\right.\iff \left\{\begin{array}{l}a=2\\ b=1\end{array}\right.$

Khi đó ta có: $\left\{\begin{array}{l}\frac{x}{x-1}=2\\ \frac{1}{y+2}=1\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=-1\end{array}\right.$

a, $\left\{\begin{array}{l}3(x+1)+2(x+2y)=4\\ 4(x+1)-(x+2y)=9\end{array}\right.$$\iff \left\{\begin{array}{l}3x+3+2x+4y=4\\ 4x+4-x-2y=9\end{array}\right.\iff \left\{\begin{array}{l}5x+4y=1\\ 3x-2y=5\end{array}\right.\iff \left\{\begin{array}{l}5x+4y=1\\ 6x-4y=10\end{array}\right.$

$\iff \left\{\begin{array}{l}11x=11\\ 6x-4y=10\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-1\end{array}\right.$

Vậy hệ đã cho có nghiệm duy nhất $\left(x;y\right)=\left(1;-1\right)$ .

b,

$\left\{\begin{array}{l}\frac{2}{x}+y=3\\ \frac{1}{x}-2y=4\end{array}\right.\iff \left\{\begin{array}{l}\frac{4}{x}+2y=6\\ \frac{1}{x}-2y=4\end{array}\right.\iff \left\{\begin{array}{l}\frac{5}{x}=10\\ \frac{1}{x}-2y=4\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{1}{2}\\ \frac{2}{x}+y=3\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{1}{2}\\ y=-1\end{array}\right.$(thỏa mãn)

Vậy hệ phương trình có nghiệm duy nhất $\left(x;y\right)=\left(\frac{1}{2};-1\right)$ .