a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

Bài 2: 

Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x< >0\\y>=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{1}{3x}+\dfrac{1}{3}\sqrt{y+3}=\dfrac{1}{4}\\\dfrac{5}{6x}+\sqrt{y+3}=\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}+\sqrt{y+3}=\dfrac{3}{4}\\\dfrac{5}{6x}+\sqrt{y+3}=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{6}\cdot\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{1}{x}+\sqrt{y+3}=\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\\\dfrac{1}{x}+\sqrt{y+3}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\\sqrt{y+3}=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y+3=\dfrac{1}{16}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{47}{16}\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x+y}}-\dfrac{2}{\sqrt{x-y}}=4\\\dfrac{2}{\sqrt{x+y}}-\dfrac{1}{\sqrt{x-y}}=5\end{matrix}\right.\)

Đặt: \(t=\sqrt{x+y}\) và \(k=\sqrt{x-y}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{t}-\dfrac{2}{k}=4\\\dfrac{2}{t}+\dfrac{1}{k}=5\end{matrix}\right.\)

Ta lại đặt: \(a=\dfrac{1}{t}\) và \(u=\dfrac{1}{k}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\2a+u=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\4a+2u=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\7a=14\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-2u=4\\a=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\a=2\end{matrix}\right.\)

Mà: 

\(u=1\Rightarrow\dfrac{1}{k}=1\Rightarrow k=1\)

\(a=2\Rightarrow\dfrac{1}{t}=2\Rightarrow t=\dfrac{1}{2}\)

Ta lại có:

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

\(t=\dfrac{1}{2}\Rightarrow\sqrt{x-y}=\dfrac{1}{2}\)

Ta có hệ:

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\2x=\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{8}-y=1\\x=\dfrac{5}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{8}\\x=\dfrac{5}{8}\end{matrix}\right.\)

Vậy \(x-\dfrac{5}{8};y=-\dfrac{3}{8}\)

Đặt 1/căn x+y=a; 1/căn x-y=b

Theo đề, ta có hệ:

3a-2b=4 và 2a+b=5

=>a=2 và b=1

=>x+y=1/4 và x-y=1

=>x=5/8 và y=-3/8

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

ĐK: \(x\ge0;y\ge3\).

Đặt \(\sqrt{x}=a;\dfrac{1}{y-3}=b\)

\(\Rightarrow\left\{{}\begin{matrix}2a+b=5\\3a-b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

Trả ẩn: \(\left\{{}\begin{matrix}\sqrt{x}=2\\\dfrac{1}{y-3}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)

Vậy hệ pt có nghiệm: \(\left(x;y\right)=\left(4;4\right)\).

a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

Đặt 1/căn x-3=a; 1/|2y-1|=b

Theo đề, ta có; 8a+11b=5 và 4a+b=3

=>a=7/9; b=-1/9

=>|2y-1|=-9(loại)

=>Hệ vô nghiệm