Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{6c}{6d}=\dfrac{3a+6c}{3b+6d}\)(1)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{7a}{7b}=\dfrac{4c}{4d}=\dfrac{7a-4c}{7b-4d}\)(2)

Từ (1) và (2) suy ra \(\dfrac{3a+6c}{3b+6d}=\dfrac{7a-4c}{7b-4d}\)

\(\Leftrightarrow\left(3a+6c\right)\left(7b-4d\right)=\left(3b+6d\right)\left(7a-4c\right)\)

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 \(\frac{a}{b}=\frac{c}{d}=k\)

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

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

b/ VT = (7a – 3b)2 – 4c2 = 49a2- 42ab + 9b2 – 4c2

mà 10a2 = 10b2 + c2 nên c2 = 10a2 – 10b2

nên VT = 49a2 – 42ab + 9b2 – 4(10a2 – 10b2)

= 49a2 – 42ab + 9b2 – 40a2 + 40b2

= 9ª2 – 42ab + 49b2 = (3a – 7b)2 = VP 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

a: \(\frac{2a+5b}{3a-4b}=\frac{2\cdot bk+5b}{3\cdot bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

\(\frac{2c+5d}{3c-4d}=\frac{2\cdot dk+5d}{3\cdot dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)

Do đó: \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

b: \(\frac{3a+7b}{5a-7b}=\frac{3\cdot bk+7b}{5\cdot bk-7b}=\frac{b\left(3k+7\right)}{b\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

\(\frac{3c+7d}{5c-7d}=\frac{3\cdot dk+7d}{5\cdot dk-7d}=\frac{d\left(3k+7\right)}{d\left(5k-7\right)}=\frac{3k+7}{5k-7}\)

Do đó: \(\frac{3a+7b}{5a-7b}=\frac{3c+7d}{5c-7d}\)

d: \(\frac{4a+9b}{4a-7b}=\frac{4\cdot bk+9b}{4\cdot bk-7b}=\frac{b\left(4k+9\right)}{b\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

\(\frac{4c+9d}{4c-7d}=\frac{4\cdot dk+9d}{4\cdot dk-7d}=\frac{d\left(4k+9\right)}{d\left(4k-7\right)}=\frac{4k+9}{4k-7}\)

Do đó: \(\frac{4a+9b}{4a-7b}=\frac{4c+9d}{4c-7d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

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