Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+b}{c+d}\\ \Leftrightarrow\left(\dfrac{a}{b}\right)^{2011}=\left(\dfrac{c}{d}\right)^{2011}=\left(\dfrac{a+b}{c+d}\right)^{2011}\\ \Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ \Rightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\\ \dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\\ \Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)
ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)
\(\Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\) (ĐPCM)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
Từ (*) ta có:
\(\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\dfrac{b^{2011}.k^{2011}+d^{2011}.k^{2011}}{b^{2011}+d^{2011}}=\dfrac{k^{2011}\left(b^{2011}+d^{2011}\right)}{b^{2011}+d^{2011}}=k^{2011}\)(1)
\(\left(\dfrac{a+c}{b+d}\right)^{2011}=\dfrac{\left(bk+dk\right)^{2011}}{\left(b+d\right)^{2011}}=\dfrac{\left[k\left(b+d\right)\right]^{2011}}{\left(b+d\right)^{2011}}=k^{2011}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)
