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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)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\Rightarrow\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}\)
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
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 \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
dựa vào nhé
Ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\Rightarrow\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{2a^4}{2b^4}=\frac{2b^4}{2c^4}=\frac{2c^4}{2d^4}=\frac{2d^4}{2e^4}\)
Áp dụng tính chất dãy tỹ số bằng nhau ta có:
\(\frac{2a^4}{2b^4}=\frac{2b^4}{2c^4}=\frac{2c^4}{2d^4}=\frac{2d^4}{2e^4}=\frac{2a^4+2b^4+2c^4+2d^4}{2b^4+2c^4+2d^4+2e^4}\)
em nghĩ là c ghi sai đề :)
Sửa lai đề : Cho a;b;c;d;e khác 0
CM : \(\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5c^4}=\frac{a}{e}\)
Giải :
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=k\)
\(\Rightarrow k^4=\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}\)
Áp dụng TC DTSBN ta được : \(k^4=\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)(1)
Ta lại có : \(k^4=k.k.k.k=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}.\frac{d}{e}=\frac{a}{e}\) (2)
Từ (1) ; (2) => \(\frac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5c^4}=\frac{a}{e}\) (đpcm)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=K\)
=> a = bK, b = cK, c = dK, d = eK
Do đó: \(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)
= \(\dfrac{2b^4K^4+3c^4K^4+4d^4K^4+5e^4K^4}{2b^4+3c^4+4d^4+5d^4}\)
= \(\dfrac{K^4\left(2b^4+3c^4+4d^4+5d^4\right)}{2b^4+3c^4+4d^4+5d^4}\)
= K4 (1)
\(\dfrac{a}{e}=\dfrac{bK}{e}=\dfrac{cK^2}{e}=\dfrac{dK^3}{e}=\dfrac{eK^4}{e}=K^4\left(2\right)\)
(1)(2) => \(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\) = \(\dfrac{a}{e}\)