a5 = 5 vi tat ca phep cchia deu = 1

\(\left\{{}\begin{matrix}\left(\dfrac{1}{13}x-7\right)^8\ge0\\\left(\dfrac{1}{17}y-7\right)^6\ge0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{13}x-7\right)^8+\left(\dfrac{1}{17}y-7\right)^6\ge0\)

Mà \(\left(\dfrac{1}{13}x-7\right)^8+\left(\dfrac{1}{17}y-7\right)^6=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{13}x-7\right)^8=0\\\left(\dfrac{1}{17}y-7\right)^6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=91\\y=119\end{matrix}\right.\)

Vậy \(x=91,y=119\)

a/3=b/8=c/5

=>2a/6=3b/24=c/5

áp dụng tc dãy ts = nhau ta có :

2a/6=3b/24=c/5=2a+3b-c/6+24-5=50/25=2

=>a/3=2=>a=6

=>b/8=2=>b=16

=>c/5=2=>c=10

=>a+b+c=6+16+10=32

10

Theo tinh chat day ti so bang nhau ta co

\(\frac{X-Y}{2-3}=\frac{-2}{-1}=2\)

Do do : X\(=\frac{X}{2}\Rightarrow X=2\cdot2=4\)

           Y = \(\frac{Y}{3}\Rightarrow Y=3\cdot2=6\)

Nho tick nha

x/2 = y/3 => x²/4 = y²/9 = k => x² = 4k ; y² = 9k

TA có: x²y² = 4k.9k = 576

36k² = 576

k² = 16 => k thuộc {-4;4}

=> x² = 4.4 = 16 mà x > 0 => x = 4

=> y² = 4.9 = 36 mà y > 0 => y = 6

\(\frac{x}{9}=\frac{3}{x^2}\)

=> \(3.9=x.x^2\)

=> 27 = \(x^3\)

=> x = 3

Em  ms có lớp thôi !