2a=3b;5b=7c và 3a+5c-7b=30
c) C =2a/3b +3b/4c +4c/5d +5d/2a biết 2a/3b = 3b/4c = 4c/5d = 5d/2a.
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 biet 2a/3b=3b/4c=4c/5d=5d/2a
ko trả lời đâu bạn ơi
C=2a/3b+3b/4c+4c/5d+5d/2a biet 2a/3b=3b/4c=4c/5d=5d/2a
Theo t/c dãy tỉ số=nhau:
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)
Khi đó \(C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=1+1+1+1=4\)
Vậy C=4
2a/3b.3b/4c.4b/5d.5d/2a biet 2a/3b=3b/4c=4c/5d=5d/2a
Tính 2a/3b + 3b/4c + 4c/5d + 5d/2a biết 2a/3b=3b/4c=4c/5d=5d/2a
2a/3b+3b/4c+4c/5d+5d/2a biết 2a/3b=3b/4c=4c/5d=5d/2a
Tính C=2a/3b+3b/4c+4c/5d+5d/2a với 2a/3b=3b/4c=4c/5d=5d/2a
Tinh
2a/3b+3b/4c+4c/5d+5d/2a
Biet 2a/3b=3b/4c=4c/5d=5d/2a
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
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 \(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=4.\)
Tính giá trị của biểu thức C=2a/3b+3b/4c+4c+5d+5d/2a biết 2a/3b=3b/4c=4c+5d=5d/2a.
tính giá trị của biểu thức sau C=2a/3b+3b/4c+4c/5d+5d/2a biết 2a/3b=3b/4c=4c/5d=5d/2a
Đặt :
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=m\)
=>\(\frac{2a}{3b}.\frac{3b}{4c}.\frac{4c}{5d}.\frac{5d}{2a}=m.m.m.m=1\)
=> m4 =1
=> m = 1
=> \(\frac{2a}{3b}=1;\frac{3b}{4c}=1;\frac{4c}{5d}=1;\frac{5d}{2a}=1\)
=>\(\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=1+1+1+1=4\)
hay C =4
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
Theo t/c dãy tỉ số=nhau:
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)
\(=>2a=3b=4c=5d\)
\(=>C=1+1+1+1=4\)
Vậy C=4