giúp mình bài này với ah.

Từ\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\Rightarrow\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}.\frac{d}{e}\)

\(\Rightarrow\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}=\frac{a}{e}\) (1)

Ta lại có : \(\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}=\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\) (TC DTSBN) (2)

Từ (1) ; (2) \(\Rightarrow\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\frac{a}{e}\) (đpcm)

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

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

Ta có : a<2b;b<3c;c<4d;d<5(a,b,c thuộc Z)

suy ra: a<2b<6c<24d<100

Để a có giá trị lớn nhất thì 2b;6c;24d cũng có giá trị lớn nhất

suy ra: 24d=96;6c=90;2b=88

Mà a<88 suy ra a=87

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\Rightarrow\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}\)

Theo TCDTSBN ta có:

\(\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}=\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(1\right)\)

Lại có: \(\frac{a^4}{b^4}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}\cdot\frac{d}{e}=\frac{a}{e}\left(2\right)\)

từ (1) và (2) => dpdcm