Question

E = c-(a+b-d)+a+(a-b) với a = \(-1\dfrac{1}{3}\);b=3\(\dfrac{1}{2}\);c=\(\dfrac{-5}{3}\);d=\(\dfrac{5}{12}\)

Answer 1

\(E=c-\left(a+b-d\right)+a+\left(a-b\right)\)

\(Với\) \(a=-1\dfrac{1}{3};b=3\dfrac{1}{2};c=\dfrac{-5}{3};d=\dfrac{5}{12}\)

\(\Rightarrow E=\dfrac{-5}{3}-\left(-1\dfrac{1}{3}+3\dfrac{1}{2}-\dfrac{5}{12}\right)+\left(-1\dfrac{1}{3}\right)+\left(-1\dfrac{1}{3}-3\dfrac{1}{2}\right)=\dfrac{-5}{3}+\dfrac{-4}{3}+\dfrac{7}{2}-\dfrac{5}{12}-\dfrac{4}{3}-\dfrac{4}{3}+\dfrac{7}{2}=\left(\dfrac{-5}{3}+\dfrac{-4}{3}-\dfrac{4}{3}-\dfrac{4}{3}\right)+\left(\dfrac{7}{2}+\dfrac{7}{2}\right)+\dfrac{5}{12}=\dfrac{-17}{3}+7+\dfrac{5}{12}=\dfrac{-68}{12}+\dfrac{84}{12}+\dfrac{5}{12}=\dfrac{21}{12}=\dfrac{7}{4}\)