ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{\sqrt{x-2}}-\frac{2}{x+y}=4\\\frac{4}{\sqrt{x-2}}-\frac{3}{x+y}+1=\frac{7}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{\sqrt{x-2}}-\frac{2}{x+y}=4\\\frac{4}{\sqrt{x-2}}-\frac{3}{x+y}=\frac{5}{2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-2}}=u>0\\\frac{1}{x+y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5u-2v=4\\4u-3v=\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=\frac{1}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

ĐK: \(x>4,y\ne-2\)

Đặt a=\(\frac{1}{\sqrt{x-4}}\left(a>0\right)\),\(b=\frac{1}{y+2}\)

Vậy \(\left\{{}\begin{matrix}\frac{3}{\sqrt{x-4}}+\frac{4}{y+2}=7\\\frac{5}{\sqrt{x-4}}-\frac{1}{y-2}=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\20a-4b=16\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}23a=23\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)(tm)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-4}}=1\\\frac{1}{y+2}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-4}=1\\y+2=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)(tm)

Vậy (x;y)=(5;-1)

a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)

ĐK: \(y\ge0,y\ne4\)

Đặt \(a=\left|x+5\right|;b=\frac{1}{\sqrt{y}-2}\left(a\ge0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-2b=4\\a+b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{10}{3}\\b=\frac{-1}{3}\end{matrix}\right.\)(TM)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+5=\frac{10}{3}\\x+5=\frac{-10}{3}\end{matrix}\right.\\\sqrt{y}-2=-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+5=\frac{10}{3}\\x+5=\frac{-10}{3}\end{matrix}\right.\\\sqrt{y}=-1\left(vl\right)\end{matrix}\right.\)

Vậy hpt vô nghiệm.

theo bài ra ta có

\(\frac{3}{\sqrt{y}-2}=-1\\ \Leftrightarrow-3=\sqrt{y}-2\\ \Leftrightarrow\sqrt{y}=-1\)

k\(^o\) có y t/m

phương trình vô ng\(^o\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+2\ge0\\2x-y\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-2\\2x\ne y\end{matrix}\right.\)

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

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

- Đặt \(a=\sqrt{x+2},b=\frac{1}{2x-y}\) ( \(a\ge0,\frac{1}{b}\ne0\) ) ta được hệ :

\(\left\{{}\begin{matrix}\frac{a}{3}+b=\frac{4}{3}\\2a+3b=5\end{matrix}\right.\)

( Đoạn này bấm máy cho nhanh nha )

=> \(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) ( TM )

- Thay lại \(a=\sqrt{x+2},b=\frac{1}{2x-y}\) ta được :

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

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

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

=> \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\) ( TM )

Vậy ...