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 

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

=>\(\frac{a^{2013}}{b^{2013}}=\frac{c^{2013}}{d^{2013}}\)=>\(\frac{2.a^{2013}}{2.b^{2013}}=\frac{5.c^{2013}}{5.d^{2013}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{2.a^{2013}}{2.b^{2013}}=\frac{5.c^{2013}}{5.d^{2013}}\)=\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\)

=>\(\frac{a^{2013}}{b^{2013}}=\frac{c^{2013}}{d^{2013}}\)=\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\)    (\(\frac{2}{2}=1;\frac{5}{5}=1\)) (1)

Áp dụng tính chất của dãy tỉ số bằng nhau :

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

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

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

Từ (1) và (2)

=>\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\)=\(\frac{\left(a+b\right)^{2013}}{\left(c+d\right)^{2013}}\)(đpcm)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{5b}=\frac{2c}{5d}\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}\Rightarrow\frac{2a^{2013}}{2c^{2013}}=\frac{5b^{2013}}{5d^{2013}}=\frac{2a^{2013}+5b^{2013}}{2c^{2013}+5d^{2013}}\)

Con hiếu bđ 7a4

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

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

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

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

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

\(\Rightarrow Q=1\)

Bài 2)

Ta có \(\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)

Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Rightarrow ab+ad< ab+bc\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)

Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)

\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Rightarrow ad+cd< bc+cd\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)

Từ (1) và (2)

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