ko trả lời đâu bạn ơi

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

nên theo tính chất dãy tỉ số băng nhau, ta có:

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

=> \(2a=3b=4c=5d\)

=> \(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2d}=1+1+1+1=4\)

Đặt \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=t.\)

                                \(\Rightarrow2a=3b.t\)

                                \(\Rightarrow3b=4c.t\)

                                \(\Rightarrow4c=5d.t\)

                                \(\Rightarrow5d=2a.t\)

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\(\Rightarrow2a+3b+4c+5d=2a.t+3b.t+4c.t+5d.t\)

\(\Rightarrow2a+3b+4c+5d=t.\left(2a+3b+4c+5d\right)\)

\(\Rightarrow t=1\)

Khi đó : \(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=t+t+t+t=1+1+1+1=4.\)

Vậy \(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4.\)

Ta có: \(C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\)

\(=\dfrac{2a}{3b}\cdot4=\dfrac{8a}{3b}\)

c) Vì  \(\dfrac{2a}{3b}=\dfrac{3b}{4c}=\dfrac{4c}{5d}=\dfrac{5d}{2a}\) nên theo t/c của DTSBN ta có :

\(\Rightarrow\)\(C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\) = \(\dfrac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

2a/3b = 3b/4c = 4c/5d = 5d/2a (1) 
ta có: 2a/3b=3b/4c=> 8ac=9b^2 
4c/5d=5d/2a=> 8ac=25d^2 
=> 9b^2=25d^2 
=> b=5d/3 
=> 3b=5d(*) 
lại có: 3b/4c=4c/5d => 3b/4c=4c/3b (theo *) 
=> 9b^2=16c^2 
=> b=4c/3 
=> 3b/4c=1 
BT= 4*3b/4c (Vì các phân số = nhau) 
=> BT=3b/c 
Mà: 3b=4c ( Vì 3b/4c=1) 
=> BT=4c/c=4 
Vậy biểu thức trên = 4

Ta có:\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

Suy ra:\(\frac{2a}{3b}=1;\frac{3b}{4c}=1;\frac{4c}{5d}=1;\frac{5d}{2a}=1\)

Thay vào C ta được:\(C=1+1+1+1=4\)

2a/3b+3b/4c+4c/5d+5d/2a biet 2a/3b=3b/4c=4c/5d=5d/2a

=>\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=\frac{2ax3bx4cx5d}{2ax3bx4cx5b}=1\)

=>1--1,4=-4

=> -4 hoac 4