ta có :

ts của a=tử số của b

mà ms của a<ms của b

suy ra a>b

sai bét

ANH HUY

a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)

\(=1.\frac{7}{12}:\frac{7}{12}\)

\(=1\)

b. 

\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)

\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)

\(=\frac{5}{9}.1\)

\(=\frac{5}{9}\)

Tk mk nha!

b)  \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)

\(=\frac{5}{9}.1\)

\(=\frac{5}{9}\)

a,\(=\frac{2}{3}+\left(\frac{-4}{27}+\frac{5}{18}\right):\frac{7}{12}\)=\(\frac{2}{3}+\frac{-16}{63}+\frac{10}{21}\)=\(\frac{8}{9}\)

Ta có : 

\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)

\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)

\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)

\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)

\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\) 

\(\Rightarrow\)\(S>10\) 

Vậy \(S>10\)

Chúc bạn học tốt ~ 

nhờ các bn trả lời nhanh dùm mk nhé! yêu các bn nhìu

\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\\ \Leftrightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}\\ \Leftrightarrow A=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\dfrac{11}{13}\\ \Leftrightarrow A=0-0+0-0+0-\dfrac{11}{13}\\ \Leftrightarrow A=\dfrac{-11}{13}\)

\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)

\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)

\(=\frac{7}{12}+\frac{31}{12}\)

\(=\frac{38}{12}=\frac{19}{6}\)

\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)

\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)

\(=\frac{-5}{9}\cdot\frac{2}{13}\)

\(=\frac{-10}{117}\)

\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)

\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)

\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)

\(=\frac{4}{5}\cdot\frac{13}{7}\)

\(=\frac{52}{35}\)

a)7/12.6/11+7/12.5/11-2.7/12

=7/12(6/11+5/11-2)

=7/12(1-2)

=7/12.(-1)

=-7/12