a) \(2xy+2x-y=8\)

\(\Rightarrow\ 2x\left(y+1\right)-\left(y+1\right)=7\)

\(\Leftrightarrow\left(2x-1\right)\left(y+1\right)=7\)

\(\Rightarrow\left[\begin{matrix}\begin{cases}2x-1=-7\\y+1=-1\end{cases}\\\begin{cases}2x-1=-1\\y+1=-7\end{cases}\end{matrix}\right.\left[\begin{matrix}\begin{cases}2x-1=7\\y+1=1\end{cases}\\\begin{cases}2x-1=1\\y+1=7\end{cases}\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=4\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=1\\y=6\end{cases}\\\left[\begin{matrix}\begin{cases}x=-3\\y=-2\end{cases}\\\begin{cases}x=0\\y=-8\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

c)\(x^2+xy+x+y=2\)

\(\Leftrightarrow x\left(x+1\right)+y\left(x+1\right)=2\)

\(\Leftrightarrow\left(x+y\right)\left(x+1\right)=2\)

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

d)\(2x^2+7xy+6y^2=15\)

\(\Leftrightarrow\left(2x+3y\right)\left(x+2y\right)=15\)

\(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\left[\begin{matrix}\begin{cases}2x+3y=1\\x+2y=15\end{cases}\\\begin{cases}2x+3y=15\\x+2y=1\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}2x+3y=3\\x+2y=5\end{cases}\\\begin{cases}2x+3y=5\\x+2y=3\end{cases}\end{matrix}\right.\end{matrix}\right.\\\left[\begin{matrix}\left[\begin{matrix}\begin{cases}2x+3y=-1\\x+2y=-15\end{cases}\\\begin{cases}2x+3y=-15\\x+2y=-1\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}2x+3y=-3\\x+2y=-5\end{cases}\\\begin{cases}2x+3y=-5\\x+2y=-3\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=-43\\y=29\end{cases}\\\begin{cases}x=27\\y=-13\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=-9\\y=7\end{cases}\\\begin{cases}x=1\\y=1\end{cases}\end{matrix}\right.\end{matrix}\right.\\\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=43\\y=-29\end{cases}\\\begin{cases}x=-27\\y=13\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=9\\y=-7\end{cases}\\\begin{cases}x=-1\\y=-1\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

\(\frac{x}{y}=\frac{5}{7}=\frac{x}{7}=\frac{y}{5}\) và x + y = 4,08

Áp dụng tính chất dãy tỉ số bằng nhau,ta có: 

   \(\frac{x}{7}=\frac{y}{5}=\frac{x+y}{7+5}=\frac{4,08}{12}=\frac{17}{50}\)

\(\frac{x}{7}=\frac{17}{50}\Rightarrow x=\frac{17.7}{50}=\frac{119}{50}\)

\(\frac{y}{5}=\frac{17}{50}\Rightarrow y=\frac{17.5}{50}=\frac{17}{10}\)

Vậy..

Còn 2 cách kia là j??? 

a, \(\frac{x}{y}=\frac{5}{7}\)và x+y=4,08

Ta có: 4,08=\(\frac{102}{25}\)

 \(\frac{x}{y}=\frac{5}{7}\Rightarrow7x=5y\)

\(\Rightarrow\frac{x}{5}=\frac{y}{7}\)và x+y=\(\frac{102}{25}\)

theo t/c dãy tỉ số bằng nhau ta có:

\(\frac{x}{5}=\frac{y}{7}=\frac{x+y}{5+7}=\frac{\frac{102}{25}}{12}=\frac{17}{50}\)

\(\Rightarrow\frac{x}{5}=\frac{17}{50}\Rightarrow x=\frac{17}{10}\)

\(\frac{y}{7}=\frac{17}{50}\Rightarrow y=\frac{119}{50}\)

vậy x=

      y=

b, \(\frac{x}{y}=\frac{3}{5}\) và x.y=1215

Ta có: \(\frac{x}{y}=\frac{3}{5}\Rightarrow5x=3y\Rightarrow\frac{x}{3}=\frac{y}{5}\)

Đặt \(\frac{x}{3}=\frac{y}{5}=k\)

=> x=3k

     y=5k

Ta có:

x.y=3k.5k=1215

hay 15k²=1215

=> k²= 81

=> k=9

=> +) \(\frac{x}{3}=9\Rightarrow x=27\)

+) \(\frac{y}{5}=9\Rightarrow y=45\)

vậy x=

      y=

-------

mình chỉ biết giải 1 cách thôi. đúng k nhá

câu 1 a) xy=-5 => (x,y)=(1,-5),(-1,5)  

b) xy=-5 với x>y=>x=1,y=-5

c)(x+1)(y-2)=-5 => * x+1=1 và y-2=-5  => x=-1, y=-3

                              * x+1=-5 và y-2=1=> x=-6 , y=3

câu 2 , câu 3 tương tự

\(3-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=2\)

\(3-\left(\frac{43}{8}+x-\frac{173}{24}\right)=2×\frac{50}{3}\)

\(3-\left(\frac{43}{8}+x-\frac{173}{24}\right)=\frac{100}{3}\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right)=3-\frac{100}{3}\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right)=-\frac{91}{3}\)

\(\frac{43}{8}+x=-\frac{91}{3}+\frac{173}{24}\)

\(\frac{43}{8}+x=-\frac{185}{8}\)

\(x=-\frac{185}{8}-\frac{43}{8}\)

\(x=-\frac{57}{2}\)

vậy \(x=-\frac{57}{2}\)

nhầm 

\(3-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=2\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=3-2\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=1\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right)=1×\frac{50}{3}\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right)=\frac{50}{3}\)

\(\frac{43}{8}+x=\frac{50}{3}+\frac{173}{24}\)

\(\frac{43}{8}+x=\frac{191}{8}\)

\(x=\frac{191}{8}-\frac{43}{8}\)

\(x=\frac{37}{2}\)

vậy \(x=\frac{37}{2}\)

Gọi\(\frac{x}{5}=\frac{y}{3}=k\)\(\Rightarrow x=5k;y=3k\)\(\Rightarrow x\times y=5k\times3k=5\times k\times3\times k=60\)

           \(\Rightarrow15k^2=60\)  \(\Rightarrow k^2=60\div15\)\(\Rightarrow k^2=4\)\(\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Với \(k=2\)

\(\Rightarrow x=10\)\(y=6\)

Với\(k=-2\)

\(\Rightarrow x=-10\)\(y=-6\)

Bài 2: 

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

Ta có: xy=12

\(\Leftrightarrow12k^2=12\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)

x³+y³=(x+y)(x²-xy+y²)=(x+y(x²+2xy+y²-3xy)=(x+y)[(x+y)²-3xy]=4[16-6]=40

Đáp số: 40

\(x+y=4\Rightarrow\left(x+y\right)^2=4^2\Leftrightarrow x^2+2xy+y^2=16\Leftrightarrow x^2+2.2+y^2=16\)

\(\Leftrightarrow x^2+4+y^2=16\Leftrightarrow x^2+y^2=12\)

=>\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=4\left(12-2\right)=4.10=40\)

\(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)

                    =\(x^3+y^3+24=64=>x^3+y^3=40\)