Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dãy tỉ số bằng nhau :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=> \(\left(\dfrac{a}{c}\right)^3=\left(\dfrac{b}{d}\right)^3=\left(\dfrac{a+b}{c+d}\right)^3\) (1)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\left(\dfrac{a}{c}\right)^3=\left(\dfrac{b}{d}\right)^3=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\) (2)
Từ (1) và (2) => \(\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\) (ĐPCM)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^3\) = \(\left(\dfrac{bk+b}{dk+d}\right)^3\) = \(\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^3\) = \(\left(\dfrac{b}{d}\right)^3\) (1)
\(\dfrac{a^3+b^3}{c^3+d^3}\) = \(\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}\) = \(\dfrac{b^3.k^3+b^3}{d^3.k^3+d^3}\) = \(\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}\) = \(\dfrac{b^3}{d^3}=\left(\dfrac{b}{d}\right)^3\) (2)
_Từ (1) và (2) suy ra:
\(\left(\dfrac{a+b}{c+d}\right)^3\) = \(\dfrac{a^3+b^3}{c^3+d^3}\)
