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

thử suy nghĩ nào

\(\frac{a}{b}=\frac{c}{d}\) 

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

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

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

=> Đpcm

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

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

=>\(\frac{a}{b}.\frac{b}{c}.\frac{c}{d}.\frac{d}{e}=\left(\frac{a+b+c+d+e}{b+c+d+e}\right)^4\)

=>\(\frac{a.b.c.d}{b.c.d.e}=\left(\frac{a+b+c+d}{b+c+d+e}\right)^4\)

=>\(\frac{a}{e}=\left(\frac{a+b+c+d}{b+c+d+e}\right)^4\)

=>đpcm

(a + 2c)(b + d) = (a + c)(b + 2d)

a(b + d) + 2c(b + d) = a(b + 2d) + c(b + 2d)

ab + ad + 2cb + 2cd = ab + 2ad + cb + 2cd

ab + ad + 2cb = ab + 2ad + cb

ad + 2cb = 2ad + cb

2ad - ad = 2cd - cd

ad = cb 

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

đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

a)

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

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

=>\(\frac{5a+2c}{5b+2d}=\frac{a-4c}{b-4d}=k\)(đpcm)

b)

\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}=\frac{b}{d}\)

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

=>\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{7k-4}{3k+5}\)

\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{7k-4}{3k+5}\)

Do đó: \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)

b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

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

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