Question

Cho x/z=z/y. Chứng minh rằng (x^2+z^2)/(y^2+z^2)=x/y

 

Answer 1

Đặt \(\frac{x}{z}=\frac{z}{y}=k\)

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

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

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

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

Answer 2

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

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

Answer 3

\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}(y2+z2x2+z2​)=yx​

\frac{x}{z}=\frac{z}{y}\Rightarrow\left(\frac{x}{z}\right)^2=\left(\frac{z}{y}\right)^2zx​=yz​⇒(zx​)2=(yz​)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)}(1)(zx​)2=(yz​)2=(z2+y2)(x2+z2)​

vì \left(2\right)\frac{x}{z}=\frac{z}{y}\Rightarrow\frac{x}{y}=\frac{z}{z}(2)zx​=yz​⇒yx​=zz​

từ (1) và (2) =>\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}(y2+z2x2+z2​)=yx​