a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4

=>-2x+y=4 và 20x+3y=2

=>x=-5/13; y=42/13

b: =>4x+2|y|=8 và 4x-3y=1

=>2|y|-3y=7 và 4x-3y=1

TH1: y>=0

=>2y-3y=7 và 4x-3y=1

=>-y=7 và 4x-3y=1

=>y=-7(loại)

TH2: y<0

=>-2y-3y=7 và 4x-3y=1

=>y=-7/5; 4x=1+3y=1-21/5=-16/5

=>x=-4/5; y=-7/5

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

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

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

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

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

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

a: =>2/x+2/y=2 và 4/x-2/y=1

=>6/x=3 và 1/x+1/y=1

=>x=2 và 1/y=1-1/2=1/2

=>x=2; y=2

b: Đặt 1/x=a; 1/y=b

=>1/3a+1/3b=1/4 và 5/6a+b=2/3

=>a=1/2; b=1/4

=>x=2; y=4

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

\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)

Lấy (2)-(1) ta đc:

\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)

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

Với \(x\ne1;y\ne-2\)

hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)

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

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

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

a: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)

b: ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)

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

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=4y4yy+12=3⇔{x=144y=36

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) = \(\left\{-\dfrac{7}{2};-\dfrac{4}{9}\right\}\)

\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)

Thay vào (1) ta được :

\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}

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

ĐK xác định : x≠0 ; y≠0

Đặt ẩn phụ : a = 1/x ; b = 1/y

Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{4};-3\) )}

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

ĐK xác định : x≠0 ; y ≠0

Áp dụng quy tác cộng đại số ta có :

\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}

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

ĐK xác định : x≠0 ; y≠0

áp dụng quy tắc cộng đại số ta có :

\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)

Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}

e) ĐK xác định x≠0 ; y≠0

\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)

Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}

Đk: \(y\ne0\)

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

\(\Rightarrow-\left(x+\dfrac{1}{y}\right)^2+\dfrac{x}{y}=\dfrac{x}{y}-2\left(x+\dfrac{1}{y}\right)\)

\(\Leftrightarrow-\left(x+\dfrac{1}{y}\right)^2+2\left(x+\dfrac{1}{y}\right)=0\)

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

TH1: \(x+\dfrac{1}{y}=0\Leftrightarrow\dfrac{1}{y}=-x\) thay vào pt dưới ta được:

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

TH2: \(x+\dfrac{1}{y}=2\Leftrightarrow\dfrac{1}{y}=2-x\) thay vào pt dưới ta được:

\(\left(2-x\right)x-2.2=-1\)\(\Leftrightarrow x^2-2x+3=0\left(vn\right)\) 

Vậy (x;y)=(-1;1);(1;-1)

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

Đem \(\left(1\right)+\left(2\right):\left(x+\dfrac{1}{y}\right)^2-2\left(x+\dfrac{1}{y}\right)=0\)

đến đây chắc bạn có thể tự làm được