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`a,`
`31/23-(7/32+8/23)`
`=31/23-7/32-8/23`
`=(31/23-8/23)-7/32`
`=1-7/32=25/32`
`b,`
`38/45-(8/45-17/51-3/11)`
`=38/45-8/45+17/51+3/11`
`= (38/45-8/45)+17/51+3/11`
`=2/3+17/51+3/11`
`=1+3/11=14/11`
`c,`
`(1/3+12/67+13/41)-(79/67-28/41)`
`= 1/3+12/67+13/41-79/67+28/41`
`= 1/3+(12/67-79/67)+(13/41+28/41)`
`= 1/3+(-1)+1=1/3+(-1+1)=1/3+0=1/3`
`d,`
`1/5+(-1/6)+1/7+(-1/8)+1/9+1/8+(-1/7)+1/6+(-1/5)`
`= (1/5+ -1/5)+(-1/6+1/6)+(1/7+ -1/7)+(-1/8 +1/8)+1/9`
`= 0+0+0+0+1/9=1/9 .`
a)\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)=\(7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)=\(\left(7\frac{4}{9}-3\frac{4}{9}\right)+4\frac{7}{11}\)= 4+\(4\frac{7}{11}\)=\(8\frac{7}{11}\)
b)\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)=\(\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5+\frac{7}{9}\)=\(\frac{-7}{9}.1+5+\frac{7}{9}\)=\(\frac{-7}{9}+\frac{7}{9}+5\)=\(\left(\frac{-7}{9}+\frac{7}{9}\right)+5\)= 0+5=5
c)\(50\%.1\frac{1}{3}.10\frac{7}{35}.0,75\)= \(\frac{1}{2}.\frac{4}{3}.10\frac{1}{5}.\frac{3}{4}\)=\(\frac{1}{2}.\frac{4}{3}.\frac{51}{5}.\frac{3}{4}\)=\(\frac{1.4.51.3}{2.3.5.4}\)=\(\frac{51}{2.5}\)=\(\frac{51}{10}\)
d)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)
a) \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=\left(\frac{67}{9}+\frac{51}{11}\right)-\frac{31}{9}\)
\(=\left(\frac{67}{9}-\frac{31}{9}\right)+\frac{51}{11}\)
\(=\frac{36}{9}+\frac{51}{11}\)
\(=\frac{95}{11}=8\frac{7}{11}\)
b) \(-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(=-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+\frac{52}{9}\)
\(=-\frac{7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=-\frac{7}{9}.1+\frac{52}{9}\)
\(=-\frac{7}{9}+\frac{52}{9}\)
= 5
d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=1.\left(1-\frac{1}{43}\right)\)
\(=1.\frac{42}{43}\)
\(=\frac{42}{43}\)
a. -37 + 54 + -70+ -163 + 246
= ( 54 + 246) + (-37 - 163 ) - 70
= 300 -200 - 70 = 30
b. -359+ 181+ -123+ 350+ -172
=(-178)+227+(-172)
=49+(-172)
=-123
c.-69+ 53+ 46+ -94+-14+ 78
=(-16)+(-48)+64
=-64+64
=0
d. 13- 12+ 11+10- 9+ 8- 7- 6+ 5- 4+ 3+2- 1
= 13 - (13 - 1) + (13 - 2) + (13 - 3) - (13 - 4) + (13 - 5) - (13 - 6) - (13 - 7) + (13 - 8) - (13 - 9) + (13 - 10) + (13 - 11) - (13 - 12)
= 13 - 13 + 1 + 13 - 2 + 13 - 3 - 13 + 4 + 13 - 5 - 13 + 6 - 13 + 7 + 13 - 8 - 13 + 9 + 13 - 10 + 13 - 11 - 13 + 12
= (13 - 13 + 13 + 13 - 13 + 13 - 13 - 13 + 13 - 13 + 13 + 13 - 13) + (1 - 2 - 3 + 4 - 5 + 6 + 7 - 8 + 9 - 10 - 11 = 12)
= 13 + 0
= 13
a) \(-37+54+\left(-70\right)+\left(-163\right)+246\)
\(=\left(246+54\right)-\left(37+163\right)-70\)
\(=300-200-70\)
\(=100-70=30\)
b) \(-359+181+\left(-123\right)+350+\left(-172\right)\)
\(=-359+181-123+350-172\)
\(=\left(-359+350\right)+\left(181-172\right)-123\)
\(=-9+9-123\)
\(=-123\)
c) \(-69+53+46+\left(-94\right)+\left(-14\right)+78\)
\(=-69+53+46-94-14+78\)
\(=\left(-69+78\right)+\left(53-14\right)+\left(46-94\right)\)
\(=9+39-48\)
\(=48-48=0\)
d) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=\left(13-12\right)+\left(11+10\right)-\left(9-8\right)-\left(7+6\right)+\left(5-4\right)+\left(3+2-1\right)\)
\(=1+21-1-13+1+5\)
\(=21-13+1+5\)
\(=8+1+5=9+5=14\)
a: =-5/11-6/11+1=-11/11+1=0
b: =-13/17-13/21-4/17=-1-13/21=-34/21
b: \(=-\dfrac{5}{12}\cdot\dfrac{9}{20}\cdot\dfrac{7}{17}=\dfrac{-21}{272}\)
d: \(=\dfrac{13}{17}\left(-\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{13}{17}\cdot\dfrac{-31}{20}=\dfrac{-403}{340}\)
a) A = 1 + 2 + 3 + 4+... + 50;
Tổng A có 50 số hạng nên A = (1 + 50).50:2 = 1275,
b) B = 2 + 4 + 6 + 8 + ...+100;
Số số hạng của tổng B là: (100 - 2): 2+1 = 50 (số)
Do đó B = (2 +100).50 : 2 = 2550.
c) C = 1 + 3 + 5 + 7 +... + 99;
Số số hạng của tổng C là: (99 - 1): 2 +1 = 50 (số)
Do đó C = (1 + 99). 50 : 2 = 2500.
d = 2 + 5 + 8 + 11 .... 98
= ( 92 - 2 ) : 3 + 1 = 33
= 33 . ( 98 + 2 ) : 2
= 1650
tick cho tớ với
a: \(=\dfrac{4\cdot11\cdot3}{4\cdot9}\cdot\dfrac{1}{121}=\dfrac{11}{121}\cdot\dfrac{1}{3}=\dfrac{1}{33}\)
b: \(=\dfrac{7}{9}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{7}{9}\)
c: \(=\dfrac{-6}{7}-\dfrac{1}{7}+\dfrac{2}{11}+\dfrac{9}{11}+\dfrac{12}{17}=\dfrac{12}{17}\)