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

\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)

\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)

Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)

Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

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

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

2bd=c(b+d)

<=>(a+c)d=bc+cd

<=>ad+cd=bc+cd

<=>ad=bc

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

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

Đề sai rồi thì phải ak

\(\left(a+c-2b\right)^{2020}+\left|2bd-cd-cb\right|^{2019}=0\) nhé !

\(\Leftrightarrow a+c-2b=0;2bd-cd-cb=0\)

\(\Leftrightarrow a+c=2b;2bd-cd-cb=0\)

\(\Leftrightarrow\left(a+c\right)d-cd-cb=0\)

\(\Leftrightarrow ad=cb\)

\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)  ( đpcm )