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1 tháng 8 2016

a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{7a}{7b}=\frac{5c}{5d}\)

\(\Rightarrow\frac{a}{b}=\frac{7a+5c}{7b+5d}\)

11 tháng 7 2017

Cho ab =cd 

Chứng minh ab =7a+5c7b+5d (7b+5d khác 0)

1 tháng 7 2018

Đặt a/b=c/d=k => a=bk,c=dk

Ta có: \(\frac{a}{b}=\frac{bk}{b}=k\left(1\right)\)

\(\frac{7a+5c}{7b+5d}=\frac{7bk+5dk}{7b+5d}=\frac{k\left(7b+5d\right)}{7b+5d}=k\left(2\right)\)

Từ (1) vavf (2) => a/b=7a+5c/7b+5d

1 tháng 7 2018

Cảm ơn bn nhìu nha

23 tháng 7 2017

\(\dfrac{a}{b}=\dfrac{7a}{7b}\\ \dfrac{c}{d}=\dfrac{5c}{5d}\Rightarrow\dfrac{a}{b}=\dfrac{7a}{7b}=\dfrac{5c}{5d}\Rightarrow\dfrac{7a}{7b}=\dfrac{5c}{5d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{7a}{7b}=\dfrac{5c}{5d}=\dfrac{7a+5c}{7b+5d}\)

\(\dfrac{7a}{7b}=\dfrac{a}{b}\Rightarrow\dfrac{a}{b}=\dfrac{5c}{5d}=\dfrac{7a+5c}{7b+5d}\Leftrightarrow\dfrac{a}{b}=\dfrac{7a+5c}{7b+5d}\)

Vậy \(\dfrac{a}{b}=\dfrac{7a+5c}{7b+5d}\left(đpcm\right)\)

23 tháng 7 2017

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

\(\Rightarrow ad=bc\)

\(\Rightarrow5ad=5bc\)

\(\Rightarrow7ab+5ad=7ab+5bc\)

\(\Rightarrow a\left(7b+5d\right)=b\left(7a+5c\right)\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{7a+5c}{7b+5d}\rightarrowđpcm\)

11 tháng 11 2023

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

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

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

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

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)