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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)
b, Có: a/b < c/d => ad < bc
Xét a.(b+d)-b.(a+c) = ab+ad-ba-bc = ad-bc < 0
=> a.(b+d) < b.(a+c)
=> a/b < a+c/b+d
c, Đề phải là cho a+b+c = 2016 chứ bạn
Có : A = a/a+b+c-c + b/a+b+c-a + c/a+b+c-b = a/a+b + b/b+c + c/c+a
Vì a,b,c thuộc Z+ nên a/a+b > 0 ; b/b+c > 0 ; c/c+a > 0
=> A > a/a+b+c + b/a+b+c + c/a+b+c = 1
Lại có : a < a+b ; b < b+c ; c < c+a => 0 < a/a+b < a ; 0 < b/b+c < 1 ; 0 < c/c+a < 1
=> A < a+c/a+b+c + b+a/a+b+c + c+b/a+b+c = 2
=> 1 < A < 2
=> A ko phải là số tự nhiên
Tk mk nha
a,ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU.
TA CÓ:\(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{d}\)=\(\frac{d}{e}\)=>\(\frac{2a^2}{2b^2}\)=\(\frac{3b^2}{3c^2}\)=\(\frac{4c^2}{4d^2}\)=\(\frac{5d^2}{5e^2}\)=\(\frac{2a^2+3b^2+4c^2+5d^2}{2b^2+3c^2+4d^2+5e^2}\)(đfcm)
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 \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}\Rightarrow\frac{a^4}{b^4}=\frac{b^4}{c^4}=\frac{c^4}{d^4}=\frac{d^4}{e^4}=\frac{2a^4}{2b^4}=\frac{3b^4}{3c^4}=\frac{4c^4}{4d^4}=\frac{5d^4}{5e^4}\)
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 \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=K\)
=> 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) => \(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\) = \(\dfrac{a}{e}\)
ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\)
\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}=\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{4d^4}{4e^4}\\ =\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}\\ \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=t\) ta có:
\(\dfrac{2a^4}{2b^4}=\dfrac{3b^4}{3c^4}=\dfrac{4c^4}{4d^4}=\dfrac{5d^4}{5e^4}=t^4\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(t^4=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)
Mặt khác: \(\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}=\dfrac{a}{e}=t.t.t.t=t^4\)
Ta có đpcm
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)