1)

\(A=\dfrac{5}{7}.\dfrac{-4}{13}+\dfrac{5}{7}.\dfrac{7}{13}-\dfrac{5}{13}.\dfrac{3}{7}\)

\(A=\dfrac{5}{7}.\dfrac{-4}{13}+\dfrac{5}{7}.\dfrac{7}{13}-\dfrac{5}{7}.\dfrac{3}{13}\)

\(A=\dfrac{5}{7}.\left(\dfrac{-4}{13}+\dfrac{7}{13}-\dfrac{3}{13}\right)\) (Tính chất phân phối của phép nhân đối với phép cộng)

\(A=\dfrac{5}{7}.0\) (Tính chất nhân với số 0)

\(A=0\)

2)

a) \(A=\dfrac{-6}{5}.\dfrac{1}{2}-\dfrac{-6}{5}.\dfrac{2}{3}+\dfrac{-6}{5}.\dfrac{3}{4}\)

\(A=\dfrac{-6}{5}.\left(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}\right)\)

\(A=\dfrac{-6}{5}.\dfrac{7}{12}\)

\(A=\dfrac{-7}{10}\)

b) \(B=\dfrac{-1}{6}.\dfrac{-3}{7}+\dfrac{4}{3}.\dfrac{-3}{7}-\dfrac{1}{2}.\dfrac{-3}{7}\)

\(B=\dfrac{-3}{7}.\left(\dfrac{-1}{6}+\dfrac{4}{3}-\dfrac{1}{2}\right)\)

\(B=\dfrac{-3}{7}.\dfrac{2}{3}\)

\(B=\dfrac{-2}{7}\)

c) \(C=\dfrac{2013}{2014}.\dfrac{5}{4}+\dfrac{2013}{2014}.\dfrac{1}{6}-\dfrac{2013}{2014}.\dfrac{17}{12}\)

\(C=\dfrac{2013}{2014}.\left(\dfrac{5}{4}+\dfrac{1}{6}-\dfrac{17}{12}\right)\)

\(C=\dfrac{2013}{2014}.0\)

\(C=0\)

a: \(A=a\left(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}\right)=\dfrac{7}{12}a=\dfrac{7}{12}\cdot\dfrac{-6}{5}=\dfrac{-7}{10}\)

b: \(B=b\left(-\dfrac{1}{6}+\dfrac{4}{3}-\dfrac{1}{2}\right)=\dfrac{2}{3}b=\dfrac{2}{3}\cdot\dfrac{-3}{7}=\dfrac{-2}{7}\)

c: \(C=c\left(\dfrac{5}{4}+\dfrac{1}{6}-\dfrac{17}{12}\right)=0\)

Thay a,b,c lần lượt vào biểu thức...

Tính được kết quả:

a) A= \(-\frac{7}{10}\)

b) B= \(-\frac{2}{7}\)

c) C= 0

a) Thay a= \(-\frac{6}{5}\)vào BT A ta có:

\(\left(-\frac{6}{5}\right).\frac{1}{2}-\left(-\frac{6}{5}\right).\frac{2}{3}+\left(-\frac{6}{5}\right).\frac{3}{4}\)= \(-\frac{7}{10}\)

Các bài dưới lần lượt thế thôi bạn

b) Thay b= \(-\frac{3}{7}\)vào BT B ta có:

\(\left(-\frac{3}{7}\right).\left(-\frac{1}{6}\right)+\left(-\frac{3}{7}\right).\frac{4}{3}-\left(-\frac{3}{7}\right).\frac{1}{2}\)= \(-\frac{2}{7}\)

c) Thay c= \(\frac{2013}{2014}\) vào BT C ta có:

\(\frac{2013}{2014}.\frac{5}{4}+\frac{2013}{2014}.\frac{1}{6}-\frac{2013}{2014}.\frac{17}{12}\)=0

C=0

A= \(\dfrac{-3}{5}-\dfrac{-4}{5}+\dfrac{-9}{10}\)

A = \(\dfrac{-7}{10}\)

đúng 100%

A=a.1/3 + a. 1/4 - a. 1/6 với a=-3/5

\(\Rightarrow\frac{-3}{5}.\frac{1}{3}+\frac{-3}{5}.\frac{1}{4}-\frac{-3}{5}.\frac{1}{6}=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=\frac{-3}{5}.\left(\frac{4}{12}+\frac{3}{12}-\frac{2}{12}\right)=\frac{-3}{5}.\frac{9}{12}=\frac{-3.9}{5.12}=\frac{-1.9}{5.4}=\frac{-9}{20}\)

B=b.5/6 + b.3/4 - b.1/2 với b = 12/13

\(\Rightarrow\frac{12}{13}.\frac{5}{6}+\frac{12}{13}.\frac{3}{4}-\frac{12}{13}.\frac{1}{2}=\frac{12}{13}.\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=\frac{12}{13}.\left(\frac{10}{12}+\frac{9}{12}-\frac{6}{12}\right)=\frac{12}{13}.1=\frac{12}{13}\)

C=c.2/3 + c. 3/4 - c. 17/12 với c = 2009/2010

\(\Rightarrow\frac{2009}{2010}.\frac{2}{3}+\frac{2009}{2010}.\frac{3}{4}-\frac{2009}{2010}.\frac{17}{12}=\frac{2009}{2010}.\left(\frac{2}{3}+\frac{3}{4}-\frac{17}{12}\right)=\frac{2009}{2010}.\left(\frac{8}{12}+\frac{9}{12}-\frac{17}{12}\right)=\frac{2009}{2010}.0=0\)