Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) (k khác 0)
➩a=bk
c=dk
Thay a=bk và c=dk vào \(\dfrac{a^2+b^2}{c^2+d^2}\) và \(\dfrac{a.b}{c.d}\)
⇒\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)
\(\dfrac{a.b}{c.d}=\dfrac{b.k.b}{d.k.d}=\dfrac{b^2}{d^2}\)
⇒\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a.b}{c.d}\) (đpcm)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2.k}{d^2.k}=\dfrac{b^2}{d^2}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\left(dpcm\right)\)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a\cdot b}{c\cdot d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a\cdot b}{c\cdot d}=\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)
