Question

a) \(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)      b) \(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{8}\)

Answer 1

a) \(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\) 

=> \(\dfrac{11}{4}\) \(-a+\dfrac{1}{4}=\dfrac{3}{2}\) 

=> \(\dfrac{11}{4}-a\) = \(\dfrac{3}{2}-\dfrac{1}{4}\) 

=> a = \(\dfrac{11}{4}-\dfrac{5}{4}\) =\(\dfrac{3}{2}\) 

Vậy a = \(\dfrac{3}{2}\) 

b) \(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{8}\) 

=> \(\dfrac{13}{4}-a-\dfrac{7}{4}=\dfrac{7}{8}\) 

=> \(\dfrac{13}{4}-a=\dfrac{21}{8}\) 

=> \(a=\dfrac{13}{4}-\dfrac{21}{8}=\dfrac{5}{8}\) 

Vậy a = \(\dfrac{5}{8}\)