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Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\) (đpcm)
Chúc bạn học tốt nha
Điều kiện nào mà bạn chứng minh được như đề bài yêu cầu đc?
3a - 4b có khác 0 không?
cậu lý ở đâu ra đấy?
Lý luận đâu?
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2a+5b}{3a-4b}=\dfrac{2bt+5b}{3bt-4b}=\dfrac{b\left(2t+5\right)}{b\left(3t-4\right)}=\dfrac{2t+5}{3t-4}\\\dfrac{2c+5d}{3c-4d}=\dfrac{2dt+5d}{3dt-4d}=\dfrac{d\left(2t+5\right)}{d\left(3t-4\right)}=\dfrac{2t+5}{3t-4}\end{matrix}\right.\Rightarrowđpcm\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+5b}{2c+5d}\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a-4b}{3c-4d}\)
\(\Rightarrow\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}=\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) \(\Rightarrow\dfrac{2bk+5b}{3bk-4b}=\dfrac{2dk+5d}{3dk-4d}\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\) Đpcm.
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
b) \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
Hình như sai sai
a. Ta có : ( a + b )( c - d ) = ac-ad+bc-bd (1)
( a - b )( c + d ) = ac+ad-bc+bd (2)
Từ giả thuyết : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\) (3)
Từ (1) , ( 2) và ( 3) \(\Rightarrow\)( a + b )( c - d) = ( a - b)( c + d )
\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Lại có :
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Theo đề ta có:
\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
=> \(\dfrac{2a+5b}{3a-4b}-\dfrac{2c+5d}{3c-4d}\)
=> \(\dfrac{a+b}{a-b}-\dfrac{c+d}{c-d}\)(1)
Mà \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)(2)
=> \(\dfrac{a-b}{c-d}\) và \(\dfrac{a+b}{c+d}\)(3)
Từ (2) và (3) => \(\dfrac{a-b}{c-d}\) = \(\dfrac{a+b}{c+d}\) = \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a-b}{c-d}\) = \(\dfrac{a+b}{c+d}\)= > \(\dfrac{a-b}{a+b}\) = \(\dfrac{c-d}{c+d}\)
=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)= \(\dfrac{a+b}{a-b}-\dfrac{c+d}{c-d}\)(4)
Từ (1) và (4)
=> \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)( đpcm)