$\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac16\\\dfrac{10}{3x}+\dfrac{10}{y}=1\\\end{cases}$

`<=>` $\begin{cases}\dfrac{10}{x}+\dfrac{10}{y}=\dfrac53\\\dfrac{10}{3x}+\dfrac{10}{y}=1\\\end{cases}$

`<=>` $\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac16\\\dfrac{20}{3}x=\dfrac23\\\end{cases}$

`<=>` $\begin{cases}x=\dfrac{1}{10}\\y=\dfrac{1}{15}\\\end{cases}$

Vậy `(x,y)=(1/10,1/15)`

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3x}+\dfrac{10}{y}=1\end{matrix}\right.\left(x,y\ne0\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3}.\dfrac{1}{x}+10.\dfrac{1}{y}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{10}{x}+\dfrac{10}{y}=\dfrac{5}{3}\left(1\right)\\\dfrac{10}{3}.\dfrac{1}{x}+\dfrac{10}{y}=1\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)-\left(2\right)\Rightarrow\dfrac{20}{3}.\dfrac{1}{x}=\dfrac{2}{3}\Rightarrow\dfrac{1}{x}=\dfrac{1}{10}\Rightarrow x=10\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\Rightarrow y=15\)

Ai giúp mình với đi ạ
Mình cảm ơn nhiều.

a) \(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)(Đk: \(x\ne-1;y\ne-1\))

Đặt \(\dfrac{x}{x+1}\)  là A

\(\dfrac{y}{y+1}\) là B 

Ta có HPT mới : \(\left\{{}\begin{matrix}2A+B=2\\A+3B=-1\end{matrix}\right.\)(1)

Giải HPT (1) ta được A=  \(\dfrac{7}{5}\) ; B=\(-\dfrac{4}{5}\)

+Với A=\(\dfrac{7}{5}\) ta có: 

\(\dfrac{x}{x+1}=\dfrac{7}{5}\)

<=>\(5x=7x+7\)

<=>-2x=7

<=> x=\(-\dfrac{7}{2}\)

+Với B = \(-\dfrac{4}{5}\) ta có:

\(\dfrac{y}{y+1}=-\dfrac{4}{5}\)

<=>5y=-4y-4

<=>9y=-4

<=>y=\(-\dfrac{4}{9}\)

Vậy HPT có nghiệm (x;y) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)

đặt \(\dfrac{1}{x+2}=a,\dfrac{1}{y+2}=b\)(\(x,y\ne-2\))

\(=>\left\{{}\begin{matrix}2a+b=1\\8a-5b=1\end{matrix}\right.=>\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{matrix}\right.\)

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

Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+2}+\dfrac{4}{y+2}=4\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{y+2}=3\\\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\end{matrix}\right.\)

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

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

Vậy: (x,y)=(1;1)

Điều kiện : x ≠ -2 ;y ≠ -2

Đặt : \(\dfrac{1}{x+2}=a;\dfrac{1}{y+2}=b\)

Ta có : 

\(hpt\text{⇔}\left\{{}\begin{matrix}2a+b=1\\8x-5b=1\end{matrix}\right.\text{⇔}\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{matrix}\right.\)

Suy ra:  

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

Vậy nghiệm của hệ phương trình : (x ; y) = (1;1)

Ta có: \(\left\{{}\begin{matrix}\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+2}+\dfrac{4}{y+2}=4\\\dfrac{8}{x+2}-\dfrac{5}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{y+2}=3\\\dfrac{2}{x+2}+\dfrac{1}{y+2}=1\end{matrix}\right.\)

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

Vậy:(x,y)=(1;1)

a) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}

b) Đk xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)

Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}

c) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)

Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}

d) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

Vậy S={(0,4;-4)}

e) ĐKXĐ : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....

Phương trình đâu bạn ?

$x=144,$ $y=36$.

a.

1x;b=1y.

⎧⎪ ⎪⎨⎪ ⎪⎩a+b=56
2416.a+15.b=215⇔{a=12b=14

⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1x=121y=13

⇔{x=2y=3
c,

xy−xy+12=1xy−12−xy=2

xy=a;xy+12=b;xy−12=c

1b+1c=2a(1)

1b−1a)+(1c−1a)=0

a−bab+a−cac=0⇔1ab−2ac=0

1a(1b−2c)=0⇒1b=2c⇒c=2b

$\Rightarrow \{\begin{array}{c}a-b=1\\ 2b-a=2\end{array}\iff \{\begin{array}{c}a=4\\ b=3\end{array}$

⇒⎧⎪ ⎪⎨⎪ ⎪⎩xy=4xy+12=3⇔⎧⎪⎨⎪⎩x=4y

b) ĐKXĐ: \(x,y\neq 0\).

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\xy=-1\end{matrix}\right.\\2y=x^3+1\end{matrix}\right.\).

Với x - y = 0 suy ra x = y. Do đó \(2x=x^3+1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1=y\left(TMĐK\right)\\x=\pm\dfrac{\sqrt{5}-1}{2}=y\left(TMĐK\right)\end{matrix}\right.\).

Với xy = -1 suy ra \(y=-\dfrac{1}{x}\). Do đó \(x^3+\dfrac{2}{x}+1=0\Rightarrow x^4+x+2=0\). Phương trình vô nghiệm do \(x^4+x+2=\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\).

Vậy...