Tính nhanh:
a, M=3/1*2+3/2*3+3/3*4+........+3/20*21
b, N=4/3*9/8*16/15*25/24*.......*100/99
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2. Tính (+5) + ( +4) = 9
3.Tính (- 9) + ( - 1) = -10
4.Tính : 28 + (-15) = 13
5.Tính (-15) + 12 = - 3
6.Tính (-25) + 25 = 0
7.Tính : 13 - 17 = - 4
8.Tính : (-78) - 9 = - 87
9.Tính : ( -24) - ( - 16) = - 8
10.Tính : 17 - (-3) = 20
11.Tính nhanh : (-21) - ( 48 + 52 - 21) = -100
2, (+5)+(+4)=+9 hoạc 9 cũng được
3, (-9)+(-1)=-10
4, 28+(-15)=13
5, (-15)+12=-3
6, (-25)+25=0
7, 13-17=4
8, (-78)-9=-87
9, (-24)-(-16)=-40
10, 17-(-3)=20
11, (-21+21)-(48-52)=-4
b)1/2*2/3*3/4*4/5*5/6*6/7*7/8*8/9*9/10
=1x2x3x4x5x6x7x8x9
2x3x4x5x6x7x8x9x10
= 1
10
a. ( 23 - 21) + ( 19 - 17) + ( 15 - 13) + ( 11 - 9) + ( 7 - 5) + ( 3 - 1)
= 2 + 2 + 2 + 2 + 2 + 2
= 2 x 6
= 12
b. ( 24 - 22 ) + ( 20 - 18 ) + ( 16 - 14 ) + ( 12 - 10) + ( 8 - 6 ) + ( 4 - 2)
= 2 + 2 + 2 + 2 + 2 + 2
= 2 x 6
= 12
a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
\(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{20.21}\)
\(M=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(M=3\left(1-\dfrac{1}{21}\right)\)
\(M=3.\dfrac{20}{21}=\dfrac{20}{7}\)
\(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{100}{99}\)
\(N=\dfrac{4.9.16.25...100}{3.8.15.24...99}\)
\(N=\dfrac{2.2.3.3.4.4.5.5...10.10}{1.3.2.4.3.5.4.6...9.11}\)
\(N=\dfrac{2.3.4.5...10}{1.2.3...9}.\dfrac{2.3.4.5...10}{3.4.5...11}\)
\(N=10.\dfrac{2}{11}=\dfrac{20}{11}\)
a) \(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+......+\dfrac{3}{20.21}\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{21}\right)\)
= \(3.\dfrac{20}{21}\)
= \(\dfrac{20}{7}\)
b) \(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.......\dfrac{100}{99}\)
= \(\dfrac{4.9.16.25.....100}{3.8.15.24.....99}\)
= \(\dfrac{2.2.3.3.4.4.5.5.......10.10}{1.3.2.4.3.5.4.6......9.11}\)
= \(\dfrac{\left(2.3.4.5.....10\right).\left(2.3.4.5.....10\right)}{\left(1.2.3.4......9\right).\left(3.4.5.....11\right)}\)
= \(\dfrac{10.2}{1.11}\)
= \(\dfrac{20}{11}\)