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NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 5 2022
1: A=1-2002/2003=1/2003
=>Nghịch đảo là 2003
2: B=179/30-59/30+3/5=4+3/5=23/5
Nghịch đảo là 5/23
DD
6 tháng 8 2015
\(A=1-\frac{2015}{2016}=\frac{1}{2016}=>\frac{1}{A}=\frac{2016}{1}\)
\(B=\frac{179}{30}-\left(\frac{59}{30}-\frac{3}{5}\right)=\frac{179}{30}-\frac{59}{30}+\frac{3}{5}=40+\frac{3}{5}=\frac{203}{5}=>\frac{1}{B}=\frac{5}{203}\)
\(C=\left(\frac{46}{5}-\frac{1}{11}\right)=\frac{46.11}{5}-1=\frac{506}{5}-1=\frac{501}{5}=>\frac{1}{C}=\frac{5}{501}\)
24 tháng 8 2017
B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004
1 x 2 x 3 x 4 x .............x 2002 x 2003
2 x 3 x 4 x .............x 2003 x 2004
1
2004
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) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)
\(C=\dfrac{501}{55}.11\)
\(C=\dfrac{501}{5}\)
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) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11=\dfrac{46}{5}\cdot11-\dfrac{1}{11}\cdot11=\dfrac{506}{5}-1=\dfrac{501}{5}\)
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)
\(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11\\ C=\left(\dfrac{506}{55}-\dfrac{5}{55}\right)\cdot11\\ \Rightarrow C=\dfrac{501}{55}\cdot11\\ \Rightarrow C=\dfrac{501}{5}\)
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
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=\(\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)
=\(\left(\dfrac{46}{5}.11\right)-\left(\dfrac{1}{11}.11\right)\)
=\(\dfrac{506}{5}-1\)
=\(\dfrac{506}{5}-\dfrac{5}{5}=\dfrac{506-5}{5}=\dfrac{501}{5}\)
câu b cậu phải phá ngoặc trước khi tính, vả lại đáp án cũng sai nữa