Ta có:

\(\left(x+y\right)^2+7\left(x+y\right)+y^2+10=0\)

\(x^2+y^2+2xy+7x+7y+y^2+10=0\)

\(x^2+y^2+1+2xy+2x+2y+5x+5y+5+4=0\)

\(\left(x+y+1\right)^2+5\left(x+y+1\right)+4=0\)

\(\left(x+y+1\right)^2+\left(x+y+1\right)+4\left(x+y+1\right)+4=0\)

\(\left(x+y+1\right)\left(x+y+2\right)+4\left(x+y+1\right)=0\)

\(\left(x+y+1\right)\left(x+y+6\right)=0\)

Max T=x+y+1=-6+1=-5 <=> x+y=-6

Min T=x+y+1=-1+1=0 <=> x+y=-1

https://hoc24.vn/cau-hoi/tim-xy-thuoc-z-thoa-man-x2-2xy-7x-y-2y2-10-0.216670050813

\(x^2+2xy+7.\left(x+y\right)+2y^2+10=0\)

\(\Leftrightarrow\left(x+y^2\right)+7.\left(x+y\right)+\dfrac{49}{4}+y^2-\dfrac{9}{4}=0\)

\(\Leftrightarrow\left(x+y+\dfrac{7}{2}^2\right)=\dfrac{9}{4}-y^2\)

\(Do\left(x+y+\dfrac{7}{2}^2\right)\ge0\Rightarrow\dfrac{9}{4}-y^2\ge0\Rightarrow y^2\le\dfrac{9}{4}\)

Mà y nguyên \(\Rightarrow\left\{{}\begin{matrix}y^2\\\\y^2=1\end{matrix}\right.=0\)

Thay vào phương trình đầu: 

Với \(y=0\Rightarrow x^2+7x+10=0\Rightarrow\left\{{}\begin{matrix}x=-2\\\\\\x=-5\end{matrix}\right.\)

Với \(y=1\Rightarrow x^2+9x+19=0\Rightarrow\) không có x nguyên

Với \(y=-1\Rightarrow x^2+5x+5=0\Rightarrow\) không có x nguyên

ta có:
1) x-5=2
x=5+2
x=7
y+7=2
Y=7-2
Y=5 
2) 3.x+2=7
3.x=7-2
3.x=5
x=5/3
1-y=7
y=1-7
y=-8
 

a,Ta có: x+y= -7/6 và y+z= 1/4

=>x+y+y+z= -7/6 +1/4

=>x+z+2y= -11/12

=>1/2+2y= -11/12

=>2y= -11/12 -1/2

=>2y= -17/12

=>y= -17/24

Mà x+y=-7/6 =>x= -7/6+17/24= -11/24

      x+z=1/2 =>z=1/2+11/24=23/24

Ta có: \(x+y=-\frac{7}{6};y+z=\frac{1}{4};x+z=\frac{1}{2}\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=-\frac{7}{6}+\frac{1}{4}+\frac{1}{2}\)

\(\Rightarrow2x+2y+2z=-\frac{28}{24}+\frac{6}{24}+\frac{12}{24}\)

\(\Rightarrow2\left(x+y+z\right)=-\frac{5}{12}\)

\(\Rightarrow x+y+z=-\frac{5}{12}:2\)

\(\Rightarrow x+y+z=-\frac{5}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(x+y\right)=-\frac{5}{24}+\frac{7}{6}\Rightarrow z=-\frac{5}{24}+\frac{28}{24}=\frac{23}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(y+z\right)=-\frac{5}{24}-\frac{1}{4}\Rightarrow x=-\frac{5}{24}-\frac{6}{24}=-\frac{11}{24}\)

\(\Rightarrow\left(x+y+z\right)-\left(x+z\right)=-\frac{5}{24}-\frac{1}{2}\Rightarrow y=-\frac{5}{24}-\frac{12}{24}=-\frac{17}{24}\)

Vậy \(x=\frac{23}{24};y=-\frac{17}{24};z=-\frac{11}{24}\)

Chuk pạn hok tốt!

b,Ta có:   x.y=1/3 và y.z= -2/5

=>x.y.y.x=1/3.(-2/5)

=>x.z.y^2= -2/15

=>-3/10.y^2= -2/15

=>y^2=4/9

=>y=2/3

  Mà x.y=1/3 =>x=1/3:2/3=1/2

       x.z= -3/10 =>z= -3/10:1/2 = -3/5

Ta co: (x+3)/(x+5) = (x+5)/(y+7)

=> (x+3).(y+7) = (x+5).(y+5)

=> xy+7x+3y+21 = xy+5x+5y+35

=> 7x-5x+21 = 5y-3y+35

=> 2x = 2y +35-21  = 2y+14

=> x = y+7

=> x-y = 7

c) tu lam nka ban!!!!

\(x^2+2xy+7\left(x+y\right)+2y^2+10=0\)0

\(< =>\left(x^2+2xy+y^2\right)+7\left(x+y\right)+y^2+10=0\)

\(< =>\left(x+y\right)^2+7\left(x+y\right)+y^2+10=0\)

Đặt a=x+y ta có

\(a^2+7a+10+y^2=0\)

\(< =>a^2+7a+\frac{49}{4}-\frac{9}{4}+y^2=0\)

\(< =>\left(a+\frac{7}{2}\right)^2+y^2=\frac{9}{4}\)

Vì \(\frac{9}{4}\)=\(0+\frac{9}{4}\)và \(a+\frac{7}{2}>=y\)nên \(\hept{\begin{cases}x+y+\frac{7}{2}=\frac{3}{2}\\y=0\end{cases}}\)\(=>\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

1: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x-y}{7-13}=\dfrac{42}{-6}=-7\)

=>x=-48; y=-91

2: x/y=3/4

=>4x=3y

=>4x-3y=0

mà 2x+y=10

nên x=3 và y=4

3: =>7x-3y=0 và x-y=-24

=>x=18 và y=42

4: =>7x-5y=0 và x+y=24

=>x=10 và y=14