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Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{3a^2+5ac}{3a^2-5ac}=\dfrac{3b^2k^2+5\cdot bk\cdot dk}{3b^2k^2-5\cdot bk\cdot dk}=\dfrac{3b^2k^2+5bdk^2}{3b^2k^2-5bdk^2}=\dfrac{3b^2+5bd}{3b^2-5bd}\)
Ta có :
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\Leftrightarrow\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7a^2-5ac}{7b^2-5bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\\ Thaya=bk;c=dk,tacó:\)
\(\dfrac{7a^2+5ac}{7b^2+5bd}=\dfrac{7\cdot b^2\cdot k^2+5\cdot bk\cdot dk}{7b^2+5bd}=\dfrac{k^2\cdot\left(7b^2+5ac\right)}{7b^2+5ac}=k^2\left(1\right)\)
\(\dfrac{7a^2-5ac}{7b^2-5bd}=\dfrac{7\cdot b^2\cdot k^2-5\cdot bk\cdot dk}{7b^2-5bd}=\dfrac{k^2\cdot\left(7b^2-5ac\right)}{7b^2-5ac}=k^2\left(2\right)\)
từ (1) và (2) \(\RightarrowĐpcm\)
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)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{a\left(7a+5c\right)}{a\left(7a-5c\right)}=\dfrac{7ck+5c}{7ck-5c}=\dfrac{c\left(7k+5\right)}{c\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(1\right)\)
\(VP=\dfrac{7b^2+5bd}{7b^2-5bd}=\dfrac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\dfrac{7dk+5d}{7dk-5d}=\dfrac{d\left(7k+5\right)}{d\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)
\(\Rightarrow\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)
Vậy \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
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)
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)\(\Rightarrow\)...............
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
