Giải:
Ta có: \(10x=6y\Rightarrow\frac{x}{6}=\frac{y}{10}\)

Đặt \(\frac{x}{6}=\frac{y}{10}=k\Rightarrow x=6k,y=10k\)

Mà \(2x^2-y^2=-28\)

\(\Rightarrow2\left(6k\right)^2-\left(10k\right)^2=-28\)

\(\Rightarrow72k^2-100k^2=-28\)

\(\Rightarrow k^2.-28=-28\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

+) \(k=1\Rightarrow x=6;y=10\)

+) \(k=-1\Rightarrow x=-6;y=-10\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(6;10\right);\left(-6;-10\right)\)

\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

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

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)

Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)

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

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

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

Chúc bạn học tốt!

10x = 6y => x = 3y/5 
thay vao ta co : 
2(3y/5)^2 - y^2 = -28 
<=> 18y^2/25 - y^2 = -28 
<=> 7y^2 = 700 
<=> y = 10 
=> x = 6 

\(10x=6y\) => \(x=\frac{6y}{10}=\frac{3y}{5}\)

=> \(2x^2-y^2=2\times\left(\frac{3y}{5}\right)^2-y^2=-28\)

<=> \(2\times\frac{9y^2}{25}-y^2=-28\)

<=> \(\frac{18y^2}{25}-y^2=-28\)

<=> \(\frac{-7y^2}{25}=-28\)

<=> \(-7y^2=-700\)

<=> \(y^2=100\)

<=> \(y=10;x=6\) hoặc \(y=-10;x=-6\)

\(10x=6y\Rightarrow\frac{x}{6}=\frac{y}{10}\Rightarrow\frac{x^2}{36}=\frac{y^2}{100}\Rightarrow\frac{2x^2}{72}=\frac{y^2}{100}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{2x^2}{72}=\frac{y^2}{100}=\frac{2x^2-y^2}{72-100}=-\frac{28}{-28}=1\)

\(\Rightarrow2x^2=72\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

\(\Rightarrow y^2=100\Rightarrow y=\pm10\)

10x=6y=>x/6=y/10=>2x^2/72=y^2/100

áp dụng tính chất dãy tỉ số bang nhau ta có

\(^{2x^2}_{72}\)=\(^{y^2}_{100}\)=2x^2-y^2/72-100=-28/-28=1

=>x=6,y=10

10x=6y suy ra y/10 = x/6 suy ra y^2/100= x^2/36 suy ra y^2/100=2.x^2/72

2x^2-y^2= -28 nên y^2 -2x^2=28

y^2/100= 2x^2/72 =( y^2 - 2x^2)/(100-72)= 28/28 =1

ý^2/100=1 suy ra y^2=100 suy ra ý=10 hoặc ý=-10

2x^2/72=1 suy ra 2x^2=72 suy ra x^2= 36 suy ra x=6 hoặc x= -6

a)\(10x=6y\Rightarrow\frac{x}{6}=\frac{y}{10}\Rightarrow\frac{x^2}{36}=\frac{2x^2}{72}=\frac{y^2}{100}\)

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x^2}{36}=\frac{2x^2}{72}=\frac{y^2}{100}=\frac{2x^2-y^2}{72-100}=\frac{-28}{-28}=1\)

\(\Rightarrow\hept{\begin{cases}x^2=1.36=36\\y^2=1.100=100\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x;y\right)=\left(-6;-10\right)\\\left(x;y\right)=\left(6;10\right)\end{cases}}\)

b)\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}\)

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x^2}{4}=\frac{y^2}{25}=\frac{x^2-y^2}{4-25}=\frac{4}{-21}\)

\(\Rightarrow\hept{\begin{cases}x^2=\frac{4}{-21}.4=-21\\y^2=\frac{4}{-21}.25=\frac{100}{-21}\end{cases}}\)

Vì \(\hept{\begin{cases}x^2\ge0\\y^2\ge0\end{cases}}\) nên ko có số x;y thỏa mãn 

Có thể bạn chép sai đề phần b rồi

thanks để mk tha

tôi chép đúng rồi

đặt \(\frac{x}{5}=\frac{y}{3}\text{ }=k\)

\(\Rightarrow\text{ }x=5k\text{ };\text{ }y=3k\)

\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Rightarrow\text{ }25k^2-9k^2=4\)

\(\Rightarrow\text{ }k^2.\left(25-9\right)=4\)

\(\Rightarrow\text{ }k^2.16=4\)

\(\Rightarrow\text{ }k^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)

\(\Rightarrow\text{ }\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)

Nếu k = \(\frac{1}{2}\)thì \(x=\frac{5}{2}\text{ };\text{ }y=\frac{3}{2}\)

Nếu k = \(-\frac{1}{2}\)thì \(x=\frac{-5}{2}\text{ };\text{ }y=\frac{-3}{2}\)

10x = 6y

\(\Rightarrow\text{ }\frac{x}{6}=\frac{y}{10}\)

đặt \(\frac{x}{6}=\frac{y}{10}=k\)

\(\Rightarrow\text{ }x=6k\text{ };\text{ }y=10k\)

\(\Rightarrow\text{ }2.\left(6k\right)^2-\left(10k\right)^2=-28\)

\(\Rightarrow\text{ }72k^2-100k^2=-28\)

\(\Rightarrow\text{ }\left(72-100\right).k^2=-28\)

\(\Rightarrow\text{ }\left(-28\right).k^2=\left(-28\right)\text{ }\)

\(\Rightarrow\text{ }k^2=\left(-28\right)\text{ }:\text{ }\left(-28\right)\)

\(\Rightarrow\text{ }k^2=1\)

\(\Rightarrow\text{ }\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

Nếu k = 1 thì x = 10 ; y = 6

Nếu k = -1 thì x = -10 ; y = -6

Ta có: \(10x=6y\) \(\Leftrightarrow x=\frac{6y}{10}=\frac{3y}{5}\)

\(\Rightarrow2x^2-y^2=2\times\left(\frac{3y}{5}\right)^2-y^2=-28\)

\(\Leftrightarrow2\times\frac{9y^2}{25}-y^2=\frac{18y^2}{25}-y^2=-28\)

\(\Leftrightarrow\frac{-7y^2}{25}=-28\Leftrightarrow-7y^2=-700\Leftrightarrow y^2=100\)

\(\Leftrightarrow x=10\) và \(y=6\) hoặc \(x=-10\) và \(y=-6\)