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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 \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
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) => \(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\) = \(\dfrac{a}{e}\)
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)
a) \(\dfrac{-12}{15}+\dfrac{-4}{26}=\dfrac{-4}{5}+\dfrac{-2}{13}=\dfrac{-52-10}{65}=\dfrac{-62}{65}\)
b) \(5\dfrac{1}{3}-2\dfrac{4}{5}=\dfrac{16}{3}-\dfrac{14}{5}=\dfrac{80}{15}-\dfrac{42}{15}=\dfrac{38}{15}\)
c) \(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)+\dfrac{-5}{10}=\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{1}{2}=\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{35}{70}=\dfrac{41}{70}\)
d) \(-1\dfrac{2}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-9}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-54}{42}+\dfrac{9}{42}-\dfrac{10}{42}=\dfrac{-55}{42}\)
e) \(12-\dfrac{11}{121}+\left(\dfrac{-8}{9}\right)-\left(-\dfrac{3}{7}\right)\)
\(=12-\dfrac{11}{121}-\dfrac{8}{9}+\dfrac{3}{7}\)
\(=\dfrac{91476}{7623}-\dfrac{693}{7623}-\dfrac{6776}{7623}+\dfrac{3267}{7623}\)
\(=\dfrac{7934}{693}\)
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\)
\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{4d^4}{4e^4}\\ =\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\\ \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
