Giải:

a) Ta có:

\(\left\{{}\begin{matrix}2a=7b\\5b=4c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{7}=\dfrac{b}{2}\\\dfrac{b}{4}=\dfrac{c}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{14}=\dfrac{b}{4}\\\dfrac{b}{4}=\dfrac{c}{5}\end{matrix}\right.\Leftrightarrow\dfrac{a}{14}=\dfrac{b}{4}=\dfrac{c}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{14}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{3a}{42}=\dfrac{7b}{28}=\dfrac{5c}{25}=\dfrac{3a+5c-7b}{42+25-28}=\dfrac{30}{39}=\dfrac{10}{13}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{14}=\dfrac{10}{13}\\\dfrac{b}{4}=\dfrac{10}{13}\\\dfrac{c}{5}=\dfrac{10}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{140}{13}\\b=\dfrac{40}{13}\\c=\dfrac{50}{13}\end{matrix}\right.\)

Vậy ...

b) Tương tự câu a.

Chúc bạn học tốt!

a,Ta có:

2a=7b\(\Rightarrow\)\(\dfrac{a}{7}\)=\(\dfrac{b}{2}\)\(\Rightarrow\)\(\dfrac{a}{14}\)=\(\dfrac{b}{4}\)(1)

5b=4c\(\Rightarrow\)\(\dfrac{b}{4}\)=\(\dfrac{c}{5}\)\(\Rightarrow\)\(\dfrac{b}{4}\)=\(\dfrac{c}{5}\)(2)

Từ (1) và (2)\(\Rightarrow\)\(\dfrac{a}{14}\)=\(\dfrac{c}{5}\)=\(\dfrac{b}{4}\)\(\Rightarrow\)\(\dfrac{3a}{42}\)=\(\dfrac{5c}{25}\)=\(\dfrac{7b}{28}\)

Áp dụng t/c dãy tỉ số bằng nhau,ta có:

\(\dfrac{3a}{42}\)=\(\dfrac{5c}{25}\)=\(\dfrac{7b}{28}\)=\(\dfrac{3a+5c-7b}{42+25-28}\)=\(\dfrac{30}{39}\)=\(\dfrac{10}{13}\)

\(\Rightarrow\)a=\(\dfrac{10}{13}\).14=\(\dfrac{140}{13}\)

b=\(\dfrac{10}{13}\).4=\(\dfrac{40}{13}\)

c=\(\dfrac{10}{13}\).5=\(\dfrac{50}{13}\)

Vậy.....

chúc bạn học tốt

b,10a=15b=21c và 3a-7b+5c=30

Từ 10a=15b=21c\(\Rightarrow\)\(\dfrac{10a}{210}\)=\(\dfrac{15b}{210}\)=\(\dfrac{21c}{210}\)hay \(\dfrac{a}{21}\)=\(\dfrac{b}{14}\)=\(\dfrac{c}{10}\)\(\Rightarrow\)\(\dfrac{3a}{63}\)=\(\dfrac{7b}{98}\)=\(\dfrac{5c}{50}\)

Áp dụng t/c dãy tỉ số bằng nhau,ta có:

\(\dfrac{3a}{63}\)=\(\dfrac{7b}{98}\)=\(\dfrac{5c}{50}\)=\(\dfrac{3a-7b+5c}{63-98+50}\)=\(\dfrac{30}{15}\)=2

\(\Rightarrow\)a=2.21=42

b=2.14=28

c=2.10=20

Vậy...

Chúc bạn học tốt

\(A=\dfrac{ab+10b+25}{ab+5a+5b+25}+\dfrac{bc+10c+25}{bc+5b+5c+25}+\dfrac{ca+10a+25}{ac+5a+5c+25}\)

\(=\dfrac{\left(ab+5b\right)+\left(5b+25\right)}{\left(ab+5a\right)+\left(5b+25\right)}+\dfrac{\left(bc+5c\right)+\left(5c+25\right)}{\left(bc+5b\right)+\left(5c+25\right)}+\dfrac{\left(ca+5a\right)+\left(5a+25\right)}{\left(ac+5a\right)+\left(5c+25\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{a\left(c+5\right)+5\left(c+5\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(=\left(\dfrac{b}{b+5}+\dfrac{5}{b+5}\right)+\left(\dfrac{a}{a+5}+\dfrac{5}{a+5}\right)+\left(\dfrac{c}{c+5}+\dfrac{5}{c+5}\right)\)

\(=1+1+1=3\) (\(a;b;c\ne-5\))

\(A=\dfrac{ab+5b+5b+25}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{bc+5c+5c+25}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{ca+5a+5a+25}{a\left(c+5\right)+5\left(c+5\right)}\)

\(A=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(A=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(A=\dfrac{a+5}{a+5}+\dfrac{b+5}{b+5}+\dfrac{c+5}{c+5}=1+1+1=3\)

Đặt 2a/5b=5b/6c=6c/7d=7d/2a=k

=> k^4=2a/5b.5b/6c.6c/7d.7d/2a=1

=>k=1 hoặc k=-1

Với k=1 thì B=4

Với k=-1 thì B=-4

Vậy B=4 hoặc B=-4

Áp dụng dãy tỉ số bằng nhau => \(\frac{2a}{5b}=\frac{5b}{6c}=\frac{6c}{7d}=\frac{7d}{2a}=\frac{2a+5b+6c+7d}{5b+6c+7d+2a}=1\)

=> \(B=1+1+1+1=4\)

Các bạn giúp ,mình gâp nhé

Các bạn ghi cả lời  giải cho mình nhé

\(\frac{2a}{5b}.\frac{5b}{6c}.\frac{6c}{7d}.\frac{7d}{2a}=1=\left(\frac{2a}{5b}\right)^4\Rightarrow\frac{2a}{5b}=1;-1\)

\(\Rightarrow B=-4;4\)

ta có :2a=5b =>a/5=b/2 hay a/15=b/6 (1)

         5b=3c=>b/3=c/5 hayb/6=c/10 (2)

từ (1) và (2) =>a/15=b/6=c/10

                       =(a+b-c) /(15+6-10)=-44/11=-4

=> a=-4*15=-60

 b=-4*6=-24

  c=-4*10=-40

(mình ko biết đúng hay sai nữa )

4a=5b=7c va a-b+c=54

\(4a=5b=7c\Rightarrow4a=5b;5b=7c\Rightarrow\frac{a}{5}=\frac{b}{4};\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{a}{5}=\frac{7b}{28};\frac{4b}{28}=\frac{c}{5}\Rightarrow\frac{a}{35}=\frac{b}{28}=\frac{c}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{a}{35}=\frac{b}{28}=\frac{c}{20}=\frac{a-b+c}{35-28+20}=\frac{54}{27}=2\)

Suy ra : \(\frac{a}{35}=2\Rightarrow a=2.35=70\)

\(\frac{b}{28}=2\Rightarrow b=2.28=56\)

\(\frac{c}{20}=2\Rightarrow c=2.20=40\)

Ta có: 4a= 5b=7c

=> \(\frac{4a}{140}=\frac{5b}{140}=\frac{7c}{140}\)

Hay:\(\frac{a}{35}=\frac{b}{28}=\frac{c}{20}\)

áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a}{35}=\frac{b}{28}=\frac{c}{20}\)=\(\frac{a-b+c}{35-28+20}=\frac{54}{27}=2\)

=>\(\frac{a}{35}=2\)

  \(\frac{b}{28}=2\)

 \(\frac{c}{20}=2\)

=> a = 70

   b = 56

  c=40

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