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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
Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)
\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)
\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)
=0
b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\cdot0=0\)
c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)
\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)
\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)
d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)
\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)
\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)
\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)
\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)
\(=0:\dfrac{3}{8}=0\)
\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)
\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)
\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)
\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)
\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)
\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)
\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)
\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)
\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
a) 5/9 + 4/9 . 3/7 + 4/9 . 4/7
= 5/9 + 4/9 . (3/7 + 4/7)
= 5/9 + 4/9 . 1
= 5/9 + 4/9
= 1
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) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=\left(13-12\right)+\left(11+10\right)-\left(9-8\right)-\left(7+6\right)+\left(5-4\right)+\left(3+2-1\right)\)
\(=1+21-1-13+1+5\)
\(=21-13+1+5\)
\(=8+1+5=9+5=14\)
1: =72/90+65/90=137/90
2: =24/56-77/56=-53/56
3: =-7/10+4/5=1/10
4: =15/100-4/100=11/100
5: =4/6-5/6=-1/6
6: =10/40-15/40-76/40=-81/40
7: =-9/10+7/18
=-81/90+35/90=-46/90=-23/45
8: =27/90-55/90=-28/90=-14/45
9: =36/60-50/60-35/60=-49/60
10: =-4/9+5/6-3/8
=-32/72+60/72-27/72
=1/72
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
ta có \(\frac{\frac{4}{5}-\frac{4}{11}+\frac{4}{13}}{\frac{9}{5}-\frac{9}{11}+\frac{9}{13}}+\frac{\frac{5}{7}+\frac{5}{11}-\frac{5}{13}}{\frac{9}{7}+\frac{9}{11}-\frac{9}{13}}\)
\(=\frac{4\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{9\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{5\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}{9\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}\)
\(=\frac{4}{9}+\frac{5}{9}=\frac{9}{9}=1\)
= 5/7.(5/11-2/11-14/4)
=5/7.(-71/22)= -355/154
A=\(\frac{5}{7}\)x\(\frac{5}{11}\)-\(\frac{5}{7}\)x\(\frac{2}{11}\)-\(\frac{5}{7}\)x\(\frac{14}{4}\)
A= \(\frac{5}{7}\)x(\(\frac{5}{11}\)-\(\frac{2}{11}\)-\(\frac{14}{4}\))
A=\(\frac{5}{7}\)x(\(\frac{3}{11}\)-\(\frac{14}{4}\))
A=\(\frac{5}{7}\)x\(\frac{-71}{22}\)
A=\(\frac{-355}{154}\)