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\(\dfrac{5a+3b}{5c+3d}=\dfrac{7a-10b}{7c-10d}\)
\(\Leftrightarrow\left(5a+3b\right)\left(7c-10d\right)=\left(5c+3d\right)\left(7a-10b\right)\)
\(\Leftrightarrow35ac-50ad+21bc-30bd=35ac-50bc+21ad-30bd\)
\(\Leftrightarrow-50ad-21ad=-50bc-21bc\)
=>-71ad=-71bc
=>ad=bc
hay a/b=c/d
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Câu hỏi của Nguyễn Thanh Huyền - Toán lớp 7 | Học trực tuyến
Ta có :\(\frac{5a+3b}{5c+3d}\)= \(\frac{7a-10b}{7c-10d}\)nên \(\frac{5a}{5c}\)=\(\frac{3b}{3d}\)=\(\frac{7a}{7c}\)=\(\frac{10b}{10d}\)(Áp dụng tính chất dãy tỉ số bằng nhau )
Do đó: \(\frac{a}{c}\)=\(\frac{b}{d}\). Hay \(\frac{a}{b}\)=\(\frac{c}{d}\)(đpcm)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2t^2+3b^2t}{11b^2t^2-8b^2}=\dfrac{b^2\left(7t^2+3t\right)}{b^2\left(11t^2-8\right)}=\dfrac{7t^2+3t}{11t^2-8}\\\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2t^2+3d^2t}{11d^2t^2-8d^2}=\dfrac{d^2\left(7t^2+3t\right)}{d^2\left(11t^2-8\right)}=\dfrac{7t^2+3t}{11t^2-8}\end{matrix}\right.\Rightarrowđpcm\)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{ab}{cd}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{7a^2+3ab}{7c^2+3cd}\)
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\)
⇒\(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Khi đó :
\(\frac{3a^2+5ab}{7a^2-10b^2}=\frac{3(bt)^2+5.bt.b}{7(bt)^2-10b^2}=\frac{b^2(3t^2+5t)}{b^2(7t^2-10)}=\frac{3t^2+5t}{7t^2-10}\)
\(\frac{3c^2+5cd}{7c^2-10d^2}=\frac{3(dt)^2+5dt.d}{7(dt)^2-10d^2}=\frac{d^2(3t^2+5t)}{d^2(7t^2-10)}=\frac{3t^2+5t}{7t^2-10}\)
\(\Rightarrow \frac{3a^2+5ab}{7a^2-10b^2}=\frac{3c^2+5cd}{7c^2-10d^2}\) (đpcm)
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) thay \(a=bk;c=dk\) ta có
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)
từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) thay \(a=bk;c=dk\) ta có
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)
\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)
\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)
từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{a.b}{c.d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a.b}{c.d}\)
\(\Rightarrow\dfrac{7a^2}{7c^2}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}=\dfrac{3a.b}{3c.d}\)
\(\Rightarrow\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
\(\Rightarrow\left(đpcm\right)\)
từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{5a+3b}{7a-2b}=\dfrac{5.ck+3.dk}{7.ck-2.dk}=\dfrac{k.\left(5c+3d\right)}{k.\left(7c-2d\right)}=\dfrac{5c+3d}{7c-2d}\)Vậy \(\dfrac{5a+3b}{7a-2b}=\dfrac{5c+3d}{7c-2d}\left(đpcm\right)\)
b) từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{7a^2+3ab}{11a^2+8.b^2}=\dfrac{7.c^2.k^2+3.c.d.k^2}{11.c^2.k^2+8.d^2.k^2}=\dfrac{k^2.\left(7.c^2+3.c.d\right)}{k^{2.}\left(11.c^2+8.d^2\right)}\) vậy .......
c)\(từ\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)
Mặt khác:\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1).(2)=>......
Từ \(\dfrac{7a-8b}{9a-10b}=\dfrac{7c-8d}{9c-10d}\)
=> \(\dfrac{7a-8b}{7c-8d}=\dfrac{9a-10b}{9c-10d}\)
Ta có : \(\dfrac{7a-8b}{7c-8d}\) = \(\dfrac{7a}{7c}=\dfrac{8b}{8d}=\dfrac{a}{c}=\dfrac{b}{d}\)
=> ad = bc (ĐPCM)