a.

ĐKXĐ: \(x\ne\pm y\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\2u+3v=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u+3v=5\\\end{matrix}\right.\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+7=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-5x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Đặt \(\left[{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\).

Ta có hệ: \(\left[{}\begin{matrix}a+b=\dfrac{1}{16}\\3a+6b=\dfrac{1}{4}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}a=\dfrac{1}{24}\\b=\dfrac{1}{48}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=24\\y=48\end{matrix}\right.\)

Vậy `(x;y)=(24;48)`.

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\(\left\{{}\begin{matrix}\dfrac{y}{x}-\dfrac{y}{x+15}=\dfrac{1}{5}\\\dfrac{-y}{x}+\dfrac{y}{x-3}=\dfrac{1}{20}\end{matrix}\right.\Leftrightarrow\)cộng đại số nhé\(\Leftrightarrow\)\(\dfrac{y}{x-3}-\dfrac{y}{x+15}=\dfrac{1}{4}\)\(\Leftrightarrow\)bạn tự qui đồng để bỏ mẫu nha\(\Leftrightarrow\)\(x^2+12x-45=72y\Leftrightarrow\)^{\(y=\dfrac{x^2=12x-45}{72}\)}

Tiếp theo bn thay cái y vừa tìm được vào 1 trong hai phương trình ở hệ trên.(thay vào y đẻ tìm x nhé , 1 trong hai cái thôi). mình thay vào cái thứ nhất nên sẽ ra là:\(\dfrac{x^2+12x-45}{72x}-\dfrac{x^2+12x-45}{72\left(x+15\right)}=\dfrac{1}{5}\). Sau đó ban qui đông giải ra sẽ tìm được x =75. Bn thay lại vào cái y vừa tìm được ở trên sẽ ra y= 90.Sau khi tìm được rồi ban nhớ hay x,y vào hệ đẻ kiểm tra lại xem nhé

Đặt \(x+1=a;y^2=b\left(b\ge0;a\ne0\right)\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{5a}+\dfrac{3b}{5}=1\\\dfrac{3}{a}+b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3a}+b=\dfrac{5}{3}\\\dfrac{3}{a}+b=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{3a}=-\dfrac{14}{3}\\\dfrac{3}{a}+b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\y=\pm\sqrt{3}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-\dfrac{3}{2};\sqrt{3}\right);\left(-\dfrac{3}{2};-\sqrt{3}\right)\)

Điều kiện xác định y>o và x>2

\(\dfrac{5}{x-2}+\dfrac{3}{y}=8\left(1\right)\)

\(\dfrac{2}{x-2}-\dfrac{3}{y}=1\left(2\right)\)

Lấy 1+2 => \(\dfrac{7}{x-2}=9=>7=9.\left(x-2\right)=>x=\dfrac{25}{9}\)(Tm)

Thay x=\(\dfrac{25}{9}\) vào 1 hoặc 2 => \(\dfrac{5}{\dfrac{25}{9}-2}+\dfrac{3}{y}=8=>y=\dfrac{21}{11}\)(TM)

Vậy.........

\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)

\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)

\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)

Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix} 14a-10b=9\\ 3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 14a-10b=9\\ 15a+10b=20\end{matrix}\right.\)

$\Rightarrow (14a-10b)+(15a+10b)=9+20$

$\Leftrightarrow 29a=29\Leftrightarrow a=1$.

$b=\frac{4-3a}{2}=\frac{1}{2}$

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

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

Đặt ẩn phụ nhé

\(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b=< =>\int_{2a-3b=1}^{a+b=3}< =>\int_{2.\left(3-b\right)-3b=1}^{,a=3-b}< =>\int_{b=1}^{a=2}\)

<=>\(\dfrac{1}{x+y}=2;\dfrac{1}{x-y}=1< =>\int_{x-y=1}^{x+y=2}< =>\int_{y=0,5}^{x=1,5}\)

Đặt :

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

Ta có hệ phương trình :

\(\left\{{}\begin{matrix}u+v=3\\2u-3v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u+2v=6\\2u-3v=1\end{matrix}\right.\)

\(\Leftrightarrow5v=5\Leftrightarrow v=1\)

Thay \(v=1\) vào phương trình thứ nhất ta đc :

\(u+1=3\Leftrightarrow u=2\)

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

\(\Leftrightarrow2y=-\dfrac{1}{2}\Rightarrow y=-\dfrac{1}{4}\)

Thay \(y=-\dfrac{1}{4}\) vào phương trình thứ 2 ta được :

\(x+\dfrac{1}{4}=1\Leftrightarrow x=\dfrac{3}{4}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}2x=3\left(x+y\right)\left(x-y\right)\\15x+3y=3\left(x+y\right)\left(x-y\right)\end{matrix}\right.\)

\(\Rightarrow2x=15x+3y\)

\(\Rightarrow15x+3y-2x=0\)

\(\Rightarrow13x+3y=0\)

\(\Rightarrow13x=-3y\Leftrightarrow x=-\dfrac{3}{13}y\)

Thay vào pt rồi tìm \(x;y\)