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21 tháng 11 2017

Ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+5b}{2c+5d}\)

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a-4b}{3c-4d}\)

\(\Rightarrow\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}=\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\left(dpcm\right)\)

21 tháng 11 2017

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 \(\left(1\right)\)\(\left(2\right)\) \(\Rightarrow\) Đpcm.

14 tháng 10 2021

a, Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

b, Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

 

 

14 tháng 10 2021

c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)

Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)

Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

AH
Akai Haruma
Giáo viên
22 tháng 8 2023

Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k$

$\Rightarrow a=bk, c=dk$

Khi đó:

$\frac{2a+3b}{3a-5b}=\frac{2bk+3b}{3bk-5b}=\frac{b(2k+3)}{b(3k-5)}=\frac{2k+3}{3k-5}(1)$

$\frac{2c+3d}{3c-5d}=\frac{2dk+3d}{3dk-5d}=\frac{d(2k+3)}{d(3k-5)}=\frac{2k+3}{3k-5}(2)$

Từ $(1); (2)$ ta có đpcm.

7 tháng 12 2017

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2a+5b}{3a-4b}=\dfrac{2bt+5b}{3bt-4b}=\dfrac{b\left(2t+5\right)}{b\left(3t-4\right)}=\dfrac{2t+5}{3t-4}\\\dfrac{2c+5d}{3c-4d}=\dfrac{2dt+5d}{3dt-4d}=\dfrac{d\left(2t+5\right)}{d\left(3t-4\right)}=\dfrac{2t+5}{3t-4}\end{matrix}\right.\Rightarrowđpcm\)

11 tháng 12 2017

\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

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 a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

c: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)

\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)

Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

1 tháng 1 2018

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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

1 tháng 1 2018

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)

\(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)(2)

=> \(\dfrac{a-b}{c-d}\)\(\dfrac{a+b}{c+d}\)(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)

15 tháng 1 2019

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 nhaok

16 tháng 1 2019

Điều kiện nào mà bạn chứng minh được như đề bài yêu cầu đc?

3a - 4b có khác 0 không?

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