Giải hệ sau :

Câu a :

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

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

Vậy ...........................

Câu b :

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)

Vậy..................

\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)

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

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

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

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

=>\(\left\{{}\begin{matrix}2x-y-3x-3y=3\\2x-6y-10-2x+y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-4y=3\\-5y=10\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y=-2\\x=-3-4y=-3-4\cdot\left(-2\right)=8-3=5\end{matrix}\right.\)

Bài 2: 

a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)

\(=5m^2-2m+9>0\forall m\)

Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m

Bài 1:

ĐKXĐ \(2x\ne y\)

Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)

HPT trở thành

\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)

`{(x+3y=x(5y-1)),(1/x-3/y=-2):}`         `ĐK: x; y ne 0`

`<=>{(x+3y=5xy-x),(-3x+y=-2xy):}`

`<=>{(5xy-2x=3y),(-3x+y=-2xy):}`

`<=>{(x(5y-2)=3y),(-3x+y=-2xy):}`

`<=>{(x=[3y]/[5y-2]),(-3x+y=-2xy):}`    `ĐK: y ne 2/5` (TH `y=2/5` ko t/m)

`<=>{(x=[3y]/[5y-2]),(-3[3y]/[5y-2]+y=-2[3y]/[5y-2]y):}`

`<=>{(x=[3y]/[5y-2]),(-9y+5y^2-2y=-6y^2):}`

`<=>{(x=[3y]/[5y-2]),(11y^2-11y=0):}`

`<=>{(x=[3y]/[5y-2]),([(y=0(ko t//m)),(y=1(t//m)):}):}`

`<=>{(x=[3. 1]/[5.1-2]=1),(y=1):}`  (t/m)

Đk: \(x\ne0,y\ne-1\)

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

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

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

\(\Rightarrow2x+3y=y+1+5\)

\(\Leftrightarrow x=3-y\) thay vào (1) có:

\(2\left(3-y\right)+3y=\left(3-y\right)y+5\)

\(\Leftrightarrow y^2-2y+1=0\)

\(\Leftrightarrow y=1\) \(\Rightarrow x=2\)(tm)

Vậy (x;y)=(2;1)

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