Question

Tính giá trị biểu thức

a, A= 1 - 2002/2003

b, B = 179/30 - ( 59/30 - 3/5)

c, C= (46/5 - 1/11) × 11

Answer 1

a) \(A=1-\dfrac{2002}{2003}\)

\(A=\dfrac{1}{2003}\)

b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\)

\(B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)-\dfrac{3}{5}\)

\(B=\dfrac{120}{30}-\dfrac{3}{5}\)

\(B=4-\dfrac{3}{5}\)

\(B=\dfrac{17}{5}\)

c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)

\(C=\dfrac{501}{55}.11\)

\(C=\dfrac{501}{5}\)

Answer 2

a) \(A=1-\dfrac{2002}{2003}=\dfrac{2003}{2003}-\dfrac{2002}{2003}=\dfrac{1}{2003}\)

b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}=4+\dfrac{3}{5}=\dfrac{23}{5}\)

c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11=\dfrac{46}{5}\cdot11-\dfrac{1}{11}\cdot11=\dfrac{506}{5}-1=\dfrac{501}{5}\)

Answer 3

a)

\(A=1-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{2003}{2003}-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{1}{2003}\)

b)

\(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\\ \Rightarrow B=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}\\ \Rightarrow B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{120}{30}+\dfrac{3}{5}\\ \Rightarrow B=4+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{20}{5}+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{23}{5}\)

c)

\(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11\\ C=\left(\dfrac{506}{55}-\dfrac{5}{55}\right)\cdot11\\ \Rightarrow C=\dfrac{501}{55}\cdot11\\ \Rightarrow C=\dfrac{501}{5}\)

Answer 4

a, A= 2003/2003 - 2002/2003 =1/2003

b, B= 179/30 - 59/30 + 3/5 = ( 179/30 - 59/30 ) + 3/5 = 120/30 + 3/5 =4 + 3/ 5 = 20/5 - 3/5 = 17/5

c, C=46/5 . 11 - 1/11 . 11 = 506/5 - 1 = 506/5 - 5/5 = 501/5

Answer 5

a, A= \(1-\dfrac{2002}{2003}\)

=\(\dfrac{2003}{2003}-\dfrac{2002}{2003}\)

=\(\dfrac{2003-2002}{2003}=\dfrac{1}{2003}\)

b, B= \(\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\)

=\(\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3.6}{5.6}\right)\)

=\(\dfrac{179}{30}-\left(\dfrac{59-18}{30}\right)\)

=\(\dfrac{179}{30}-\dfrac{41}{30}\)

=\(\dfrac{179-41}{30}=\dfrac{138}{30}=\dfrac{23}{5}\)

c, C=\(\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)

=\(\left(\dfrac{46}{5}.11\right)-\left(\dfrac{1}{11}.11\right)\)

=\(\dfrac{506}{5}-1\)

=\(\dfrac{506}{5}-\dfrac{5}{5}=\dfrac{506-5}{5}=\dfrac{501}{5}\)