Đặt: \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=k\)
=> 2a = k .3b; 3b = k. 4c; 4c = k. 5d; 5d = k.2a
Mà \(1=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=\frac{k.3b+k4c+k.5d+k.2a}{3b+4c+5d+2a}=\frac{k.\left(3b+4c+5d+2a\right)}{3b+4c+5d+2a}=k\)
=> C = 1+1+1+1 = 4
cho a,b,c,d khác 0 biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\) Tính: \(C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\)
Cho a,b,c,d \(\ne0\) biết:\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}.\text{Tính}C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\)
Tính:
\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}biết\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
C=\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\) biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
Tính giá trị biểu thức:
\(a,A=\left(-1\right).\left(-1\right)^2.\left(-1^3\right).\left(-1\right)^4...........\left(-1\right)^{2016}.\left(-1\right)^{2017}\)
\(b,B=70.\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)
\(c,C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\)
biết\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
Tính :
\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\) biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
giúp mình nha !
giúp tôi vs mi na
C = \(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\) biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
1. Cho a, b, c, d\(\ne\) 0 biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
Tính C=\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
\(C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\)biết \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}\)
