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a: 2(x+y)=5(y+z)=3(x+z)

=>\(\frac{2\left(x+y\right)}{30}=\frac{5\left(y+z\right)}{30}=\frac{3\left(x+z\right)}{30}\)

=>\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)

Đặt \(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}=k\)

=>x+y=15k; y+z=6k; x+z=10k

=>x+y-x-z=15k-10k=5k; y+z=6k

=>y-z=5k và y+z=6k

=>y=(5k+6k)/2=5,5k; z=5,5k-5k=0,5k

x+y=15k

=>x+5,5k=15k

=>x=9,5k

x-y=9,5k-5,5k=4k; y-z=5,5k-0,5k=5k

=>\(\frac{x-y}{4}=\frac{y-z}{5}\)

Đặt \(\frac{x}{z}=\frac{z}{y}=k\)

\(\Rightarrow\hept{\begin{cases}x=zk\\z=yk\end{cases}}\)

Khi đó : \(\frac{x^2+z^2}{y^2+z^2}=\frac{\left(zk\right)^2+z^2}{y^2+\left(yk\right)^2}=\frac{z^2\left(k^2+1\right)}{y^2\left(k^2+1\right)}=\frac{z^2}{y^2}=\frac{\left(y.k\right)^2}{y^2}=k^2\)

\(\frac{x}{y}=\frac{y.k^2}{y}=k^2\)

=> \(\frac{x^2+z^2}{y^2+z^2}=\frac{x}{y}\left(\text{đpcm}\right)\)

\(\frac{x}{z}=\frac{z}{y}\)

cmr: \(\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}\)

\(\frac{x}{z}=\frac{z}{y}\Rightarrow\left(\frac{x}{z}\right)^2=\left(\frac{z}{y}\right)^2\)

áp dụng t/c dãy tỉ số = nhau

\(\left(1\right)\left(\frac{x}{z}\right)^2=\left(\frac{z}{y}\right)^2=\frac{\left(x^2+z^2\right)}{\left(z^2+y^2\right)}\)

vì \(\left(2\right)\frac{x}{z}=\frac{z}{y}\Rightarrow\frac{x}{y}=\frac{z}{z}\)

từ (1) và (2) =>\(\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}\)

Ta có \(\frac{x}{z}=\frac{z}{y}\)=> \(\frac{x^2}{z^2}=\frac{z^2}{y^2}=\frac{x^2+z^2}{z^2+y^2}\)(1)

Ta lại có : \(\frac{x}{z}=\frac{z}{y}\)=> \(xy=z^2\)(2) 

Từ (1), (2) có: \(\frac{x^2+z^2}{y^2+z^2}=\frac{x^2}{z^2}=\frac{x^2}{xy}=\frac{x}{y}\)(đpcm)