\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk và c=dk
ta có \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)\(\dfrac{ab}{cd}=\dfrac{bk.b}{bk.d}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (cùng =\(\dfrac{b^2}{d^2}\) ) (đpcm)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=t\)
a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\end{matrix}\right.\Rightarrowđpcm\)
b) \(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\left(\dfrac{a+b}{c+d}\right)^2=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k:\)
Nên a=kb, c=kd
Do đó: \(\dfrac{ab}{cd}=\dfrac{kbb}{kdd}=\dfrac{kb^2}{kd^2}=\dfrac{b^2}{d^2}\)
Và: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(kb\right)^2-b^2}{\left(kd\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
Vậy \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(kb+b\right)^2}{\left(kd+d\right)^2}=\dfrac{\left[\left(k+1\right)b\right]^2}{\left[\left(k+1\right)d\right]^2}=\dfrac{\left(k+1\right)^2.b^2}{\left(k+1\right)^2.d^2}=\dfrac{b^2}{d^2} \)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\dfrac{\left(k^2+1\right)b^2}{\left(k^2+1\right)d^2}=\dfrac{b^2}{d^2}\)
Vậy \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
