đặt \(\frac{2a}{5b}=\frac{5b}{6c}=\frac{6c}{7d}=\frac{7d}{2a}=k\)

\(\frac{2a}{5b}.\frac{5b}{6c}.\frac{6c}{7d}.\frac{7d}{2a}=\frac{2a.5b.6c.7d}{5b.6c.7d.2a}=1=k^4\)

\(\Rightarrow k\in\left\{-1;1\right\}\)

\(\frac{2a}{5b}>\frac{0}{5b}=0\Rightarrow k=1\)

vậy \(A=\frac{2a}{5b}+\frac{5b}{6c}+\frac{6c}{7d}+\frac{7d}{2a}=4k=4.1=4\)

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

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

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

giúp mình nhé

\(a,\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2b+2+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1+1=2\\b=1\end{matrix}\right.\\ b,\Leftrightarrow\left\{{}\begin{matrix}2a-4+a=7\\b=4-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{11}{3}\\b=4-\dfrac{11}{3}=\dfrac{1}{3}\end{matrix}\right.\)

\(8a\left(a+2b\right)=5b\left(b+2a\right)\)

\(\Leftrightarrow8a^2+16ab=5b^2+10ab\)

\(\Leftrightarrow8a^2+6ab-5b^2=0\)

\(\Leftrightarrow\left(2a-b\right)\left(4a+5b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{b}{2}\\a=-\dfrac{5b}{4}\end{matrix}\right.\)

\(8a\left(a+2b\right)=5b\left(2a+b\right)\)

\(\Leftrightarrow8a^2+16ab-10ab-5b^2=0\)

\(\Leftrightarrow8a^2+6ab-5b^2=0\)

\(\Leftrightarrow8a^2+10ab-4ab-5b^2=0\)

\(\Leftrightarrow\left(4a+b\right)\left(2a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-\dfrac{b}{4}\\a=\dfrac{b}{2}\end{matrix}\right.\)

Ta có : 

\(2a=\frac{a}{\frac{1}{2}};3b=\frac{b}{\frac{1}{3}};5b=\frac{b}{\frac{1}{5}};7c=\frac{c}{\frac{1}{7}}\)

Lại có \(\hept{\begin{cases}\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}\\\frac{b}{\frac{1}{5}}=\frac{c}{\frac{1}{7}}\end{cases}}\Rightarrow\frac{a}{\frac{3}{2}}=b=\frac{c}{\frac{5}{7}}\Leftrightarrow\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}\)

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

\(\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}=\frac{3a-7b+5c}{\frac{9}{2}-1+\frac{25}{7}}=\frac{-30}{\frac{99}{14}}=\frac{-140}{33}\)

\(\Rightarrow\hept{\begin{cases}3a=\frac{-140}{33}\cdot\frac{9}{2}=\frac{-210}{11}\Rightarrow a=\frac{-70}{11}\\7b=\frac{-140}{33}\Rightarrow b=\frac{-20}{33}\\5c=\frac{-140}{33}\cdot\frac{25}{7}=\frac{-500}{33}\Rightarrow c=\frac{-100}{33}\end{cases}}\)

Vậy....

Chắc sai =))