Question

[GT 6916]

Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\left(b\ne d\right)\).CTR ta có tỉ lệ thức

\(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\)

Answer 1

[ST]

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a=bk,c=dk

Ta có: \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{\left(bk\right)^2+b^2}=\frac{\left[b\left(k+1\right)\right]^2}{b^2k^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(1\right)\)

\(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(dk+d\right)^2}{\left(dk\right)^2+d^2}=\frac{\left[d\left(k+1\right)\right]^2}{d^2k^2+d^2}=\frac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\left(2\right)\)

Từ (1) và (2) => đpcm