a: \(\begin{cases}\frac{5x}{3}-\frac{2y}{5}=19\\ 4x+\frac{3y}{2}=21\end{cases}\Rightarrow\begin{cases}25x-6y=285\\ 8x+3y=42\end{cases}\)
=>\(\begin{cases}25x-6y=285\\ 16x+6y=84\end{cases}\Rightarrow\begin{cases}25x-6y+16x+6y=285+84\\ 8x+3y=42\end{cases}\)
=>\(\begin{cases}41x=369\\ 3y=42-8x\end{cases}\Rightarrow\begin{cases}x=9\\ 3y=42-8\cdot9=42-72=-30\end{cases}\Rightarrow\begin{cases}x=9\\ y=-10\end{cases}\)
b: \(\begin{cases}3x^2+y^2=\frac{31}{36}\\ x^2-3y^2=-\frac{1}{12}\end{cases}\Rightarrow\begin{cases}3x^2+y^2=\frac{31}{36}\\ 3x^2-9y^2=-\frac14\end{cases}\)
=>\(\begin{cases}3x^2+y^2-3x^2+9y^2=\frac{31}{36}+\frac14=\frac{31}{36}+\frac{9}{36}=\frac{40}{36}\\ x^2-3y^2=-\frac{1}{12}\end{cases}\)
=>\(\begin{cases}10y^2=\frac{10}{9}\\ x^2=3y^2-\frac{1}{12}\end{cases}\Rightarrow\begin{cases}y^2=\frac19\\ x^2=3\cdot\frac19-\frac{1}{12}=\frac13-\frac{1}{12}=\frac14\end{cases}\)
=>\(\begin{cases}y\in\left\lbrace\frac13;-\frac13\right\rbrace\\ x\in\left\lbrace\frac12;-\frac12\right\rbrace\end{cases}\)
c: ĐKXĐ: x>=1; y>=1
\(\begin{cases}2\sqrt{x-1}-\sqrt{y-1}=1\\ \sqrt{x-1}+\sqrt{y-1}=2\end{cases}\Rightarrow\begin{cases}2\sqrt{x-1}-\sqrt{y-1}+\sqrt{x-1}+\sqrt{y-1}=1+2\\ \sqrt{x-1}+\sqrt{y-1}=2\end{cases}\)
=>\(\begin{cases}3\sqrt{x-1}=3\\ \sqrt{y-1}=2-\sqrt{x-1}\end{cases}\Rightarrow\begin{cases}\sqrt{x-1}=1\\ \sqrt{y-1}=2-1=1\end{cases}\Rightarrow\begin{cases}x-1=1\\ y-1=1\end{cases}\)
=>x=2(nhận) và y=2(nhận)
d: \(\begin{cases}\left(x-1\right)^2-2y=\frac{14}{9}\\ 3\left(x-1\right)^2+3y=-\frac13\end{cases}\Rightarrow\begin{cases}3\left(x-1\right)^2-6y=\frac{14}{3}\\ 3\left(x-1\right)^2+3y=-\frac13\end{cases}\)
=>\(\begin{cases}3\left(x-1\right)^2-6y-3\left(x-1\right)^2-3y=\frac{14}{3}+\frac13\\ \left(x-1\right)^2-2y=\frac{14}{9}\end{cases}\)
=>\(\begin{cases}-9y=\frac{15}{3}\\ \left(x-1\right)^2=2y+\frac{14}{9}\end{cases}\Rightarrow\begin{cases}y=-\frac{15}{27}=-\frac59\\ \left(x-1\right)^2=2\cdot\frac{-5}{9}+\frac{14}{9}=\frac49\end{cases}\)
=>\(\begin{cases}y=-\frac59\\ x-1\in\left\lbrace\frac23;-\frac23\right\rbrace\end{cases}\Rightarrow\begin{cases}y=-\frac59\\ x\in\left\lbrace\frac53;\frac13\right\rbrace\end{cases}\)
e: ĐKXĐ: x+y-1<>0 và 2x-y+3<>0
=>y<>-x+1 và y<>2x+3
\(\begin{cases}\frac{4}{x+y-1}-\frac{5}{2x-y+3}=\frac52\\ \frac{3}{x+y-1}+\frac{1}{2x-y+3}=\frac75\end{cases}\Rightarrow\begin{cases}\frac{4}{x+y-1}-\frac{5}{2x-y+3}=\frac52\\ \frac{15}{x+y-1}+\frac{5}{2x-y+3}=7\end{cases}\)
=>\(\frac{15}{x+y-1}+\frac{5}{2x-y+3}+\frac{4}{x+y-1}-\frac{5}{2x-y+3}=7+\frac52\)
=>\(\frac{19}{x+y-1}=\frac{19}{2}\)
=>x+y-1=2
=>x+y=3
\(\frac{4}{x+y-1}-\frac{5}{2x-y+3}=\frac52\)
=>\(\frac42-\frac{5}{2x-y+3}=\frac52\)
=>\(\frac{5}{2x-y+3}=\frac{-1}{2}\)
=>2x-y+3=-10
=>2x-y=-13
=>2x-y+x+y=-13+3
=>3x=-10
=>x=-10/3(nhận)
x+y=3
=>y=-3+10/3=1/3(nhận)
f: ĐKXĐ: x<>y và x<>-y
\(\begin{cases}\frac{2}{x+y}+\frac{1}{x-y}=3\\ \frac{1}{x+y}-\frac{3}{x-y}=1\end{cases}\Rightarrow\begin{cases}\frac{6}{x+y}+\frac{3}{x-y}=9\\ \frac{1}{x+y}-\frac{3}{x-y}=1\end{cases}\)
=>\(\begin{cases}\frac{6}{x+y}+\frac{3}{x-y}+\frac{1}{x+y}-\frac{3}{x-y}=9+1\\ \frac{2}{x+y}+\frac{1}{x-y}=3\end{cases}\Rightarrow\begin{cases}\frac{7}{x+y}=10\\ \frac{1}{x-y}=3-\frac{2}{x+y}\end{cases}\)
=>\(\begin{cases}x+y=\frac{7}{10}\\ \frac{1}{x-y}=3-2:\frac{7}{10}=3-\frac{20}{7}=\frac17\end{cases}\Rightarrow\begin{cases}x+y=\frac{7}{10}\\ x-y=7\end{cases}\)
=>\(\begin{cases}x=\left(\frac{7}{10}+7\right):2=\frac{77}{10}:2=\frac{77}{20}\\ y=x-7=\frac{77}{20}-7=\frac{77}{20}-\frac{140}{20}=-\frac{63}{20}\end{cases}\) (nhận)
