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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a+b}{c+d}+\frac{a+b}{c+d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2.b^2}{c^2.d^2}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b^2\right)}{\left(c+d\right)^2}\)
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\left(2\right)\)
từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2.k}{d^2,k}=\dfrac{b^2}{d^2}\)(3)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(4)
từ (3) (4) \(\Rightarrow\)......
c) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2}{d^2}\) (5)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\left(6\right)\)
từ (5) (6)\(\Rightarrow\)...............
b) Ta có:
\(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a}{c}=\frac{b}{d}\) => \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
Mặt khác \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\) => \(\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\left(\frac{a+b}{c+d}\right)^2\)
Vậy:.........
a) Ta có:
\(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a^2}{b^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\)
Theo tc dãy tỉ số bằng nhau ta có:
\(\frac{7a^2}{7b^2}=\frac{5ac}{5bd}=\frac{7a^2+5ac}{7b^2+5bd}=\frac{7a^2-5ac}{7b^2-5bd}\)
=>\(\frac{7a^2+5ac}{7a^2-5ac}=\frac{7b^2+5ad}{7b^2-5ad}\)
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Với \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{b}.\)\(\dfrac{c}{d}=\dfrac{ac}{bd}=\dfrac{aa}{bb}=\dfrac{a^2}{b^2}\)
Ta có : \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)
=> \(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{7a^2}{7b^2}=\dfrac{5ac}{5bd}=\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\) (1)
Từ (1) => \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2-5bd}{7b^2-5bd}\) (ĐPCM)
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