t.i.c.k mik mik t.i.c.k lại

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

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

\(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{3bk-4dk}{3b-4d}=\frac{k.\left(3b-4d\right)}{3b-4d}=k\)(1)

    \(\frac{5a-6c}{5b-6d}=\frac{5bk-6dk}{5b-6d}=\frac{k.\left(5b-6d\right)}{5b-6d}=k\)(2)

Từ (1) và  (2)

\(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{5a-6c}{5b-6d}\)

                                đpcm

Mà kudo shinichi này,CTV là j

CTV là Cộng Tác Viên

I don't now

or no I don't

..................

sorry

ta có: \(\frac{3a-4c}{3b-4d}=\frac{3a}{3b}=\frac{4c}{4d}=\frac{a}{b}=\frac{c}{d}\)

\(\frac{5a+6c}{5b+6d}=\frac{5a}{5b}=\frac{6c}{6d}=\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{3a-4c}{3b-4d}=\frac{5a+6c}{5b+6d}\left(=\frac{a}{b}=\frac{c}{d}\right)\)

tự làm đi

Kiu ai làm cho hả???

_Người lạ lướt web_

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)

Ta có:

\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{3a}{3b}=\frac{3c}{3d}\)=>\(\frac{3a}{3c}=\frac{3b}{3d}\)                                 ;            \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{4a}{4b}=\frac{4c}{4d}\)=>\(\frac{4a}{4c}=\frac{4b}{4d}\)

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

\(\frac{3a}{3c}=\frac{3b}{3d}=\frac{3a+3b}{3c+3d}\)                                                           ;              \(\frac{4a}{4c}=\frac{4b}{4d}=\frac{4a+4b}{4c+4d}\)

Mà \(\frac{3a}{3b}=\frac{3b}{3d}=\frac{4a}{4c}=\frac{4b}{4d}\)

=>\(\frac{3a+3b}{3c+3d}=\frac{4a+4b}{4c+4d}\)

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