\(\frac{a}{b}=\frac{c}{d}\\ \Rightarrow\frac{a}{c}=\frac{b}{d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\\ \Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}\left(1\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)

Đặt\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=k\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=k\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=k^{2013}\)(1)

Mặt khác:\(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=k^{2013}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}=k^{2013}\)(2)

Từ (1);(2) ta có: \(\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(=k^{2013}\right)\)

có \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)=>\(\frac{a^{2013}}{c^{2013}}=\frac{\left(a-b\right)^{2013}}{\left(c-d\right)^{2013}}\)

ngược lại cũng có \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)

=> đpcm :V 

Bài làm

Vì\(\frac{a}{b}=\frac{c}{d}\)

=> \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\)

=> \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)(đpcm)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

       \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}\) (1)

Ta có: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

            \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\) (2)

Từ (1) và (2), ta được: \(\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\).Thế vào,ta có:

\(\left(\frac{a-b}{c-d}\right)^{2013}=\left(\frac{kb-b}{kd-d}\right)^{2013}=\left(\frac{b\left(k-1\right)}{d\left(k-1\right)}\right)^{2013}=\frac{b^{2013}}{d^{2013}}\) (1)

\(\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}=\frac{\left(kb\right)^{2013}+b^{2013}}{\left(kd\right)^{2013}+d^{2013}}=\frac{k^{2013}b^{2013}+b^{2013}}{k^{2013}d^{2013}+d^{2013}}=\frac{b^{2013}\left(k^{2013}+1\right)}{d^{2013}\left(k^{2013}+1\right)}=\frac{b^{2013}}{d^{2013}}\) (2)

Từ (1) và (2) ta có: \(\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}^{\left(đpcm\right)}\)

Từ \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(V_1=\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}.k^{2013}}{d^{2013}\cdot k^{2013}}=\frac{b^{2013}}{d^{2013}}\)

\(V_3=\left(\frac{a-b}{c-d}\right)^{2013}=\left(\frac{bk-b}{dk-d}\right)^{2013}=\left[\frac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2013}=\frac{b^{2013}}{d^{2013}}\)

\(\Rightarrow V_1=V_2=V_3\)

Hay \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}\left(đpcm\right)\)

a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{c}{a}=\frac{d}{b}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{c}{a}=\frac{d}{b}=\frac{c+d}{a+b}\)(đpcm)

a) Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{c}{a}=\frac{d}{b}=\frac{c+d}{a+b}\Rightarrow\frac{c+d}{a+b}=\frac{c}{a}\left(\text{đpcm}\right)\)

b)Ta có : \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(\text{đpcm}\right)\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left(a^{2013}+b^{2013}\right)\left(b^{2013}+c^{2013}\right)\left(c^{2013}+a^{2013}\right)=0\)

\(\Rightarrow P=\frac{17}{25}\)

\(a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)+2abc=0\)

=>\(\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

=>a=-b hoặc a=-c hoặc b=-c (1)

=>a=1 hoăc b=1 hoặc c=1 (2)

từ 1 và 2 => Q=1

gt \(\Rightarrow\left\{{}\begin{matrix}b\left(a^2+2ac+c^2\right)+ac\left(a+c\right)+b^2\left(a+c\right)=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\\a^{2013}+b^{2013}+c^{2013}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=0\Rightarrow a^{2013}+b^{2013}=0\\b+c=0\Rightarrow b^{2013}+c^{2013}=0\\a+c=0\Rightarrow a^{2013}+c^{2013}=0\end{matrix}\right.\\a^{2013}+b^{2013}+c^{2013}=1\end{matrix}\right.\)

\(\Rightarrow Q=1\)