Bài 77 trang 39 SGK Toán 6 tập 2

Tính giá trị các biểu thức sau:

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

   \(B=\dfrac{3}{4}.b+\dfrac{4}{3}.b-\dfrac{1}{2}.b\)   với \(b=\dfrac{6}{19}\) ;

   \(C=c.\dfrac{3}{4}+c.\dfrac{5}{6}-c.\dfrac{19}{12}\)  với \(c=\dfrac{2002}{2003}\) ; 

+) Ta có: \(A=a.\left (\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4} \right )\)\(=a.\dfrac{6+4-3}{12}=a.\dfrac{7}{12}\)

Với \(a= \dfrac{-4}{5}\) , thì \(A=\dfrac{-4}{5}.\dfrac{7}{12}=\dfrac{-7}{15}.\)

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

\(=b\left( {\dfrac{3}{4} + \dfrac{4}{3} - \dfrac{1}{2}} \right) \)\(= b.\dfrac{{3.3 + 4.4 - 1.6}}{{12}} = b.\dfrac{{19}}{{12}}\)

Với \(b = \dfrac{6}{{19}} \Rightarrow B = \dfrac{6}{{19}}.\dfrac{{19}}{{12}}\)\( = \dfrac{1}{2}\)

Vậy  \(B=\dfrac{1}{2}\) 

+) \(C = c.\left( {\dfrac{3}{4} + \dfrac{5}{6} - \dfrac{{19}}{{12}}} \right) \)

\(= c.\left( {\dfrac{{3.3}}{{12}} + \dfrac{{5.2}}{{12}} - \dfrac{{19}}{{12}}} \right) \)\(= c.0 = 0\)

Hay \(C = 0.\)