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Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)
Vậy \(A< \frac{3}{4}\)
A=1+32+34+.....+32008
32A-A=1+32+34+....+32008+32010-[1+32+34+...+32008]
9A-A=32010
8A=32010
Mình làm vậy đúng hay sai.
A=1+32+33+34+......+32008
3A=3+33+34+35+......+32009
3A-A=(3+33+34+35.....+32009)-(1+32+33+34+...+32008)
A=(3+32009)-(1+32008)=(3+31+32008)-(1+32008)=3+3-1=5
gọi biểu thức là A
Ta có
A=(-1/2):(-2/3):(-3/4):...:(-98/99):(-99/100)
A=\(\frac{\text{-1*(-3)*(-4)*...*(-99)*(-100) }}{2\cdot2\cdot3\cdot...\cdot98\cdot99}\)
A=\(\frac{1\cdot100}{2\cdot2}\)
A=25
`x/(2*2)+x/(3*4)+x/(6*4)+x/(8*5)+x/(10*6)=5`
`x/4 +x/12 +x/24 +x/45 + x/60 =5`
`x(1/4 +1/12 +1/24 +1/40 +1/60)=5`
`x( (30+10+5+3+2)/120 ) =5`
`x *50/120 =5`
`x = 5 *120/50 =12`
a) A:3 dư 1 => A = 7
B:3 dư 2 = 8
=> A nhân B = 7 nhân 8 chia 3 = 56 = 18 dư 4
b) A:9 dư 7 => A = 25
B:9 dư 4 => B = 22
=> A nhân B = 25 nhân 22 chia 9 = 550 : 9 = 61 dư 1
\(2+\dfrac{1}{2}-\dfrac{4}{3}+\left(-\dfrac{1}{3}\right)^4\)
=\(2+\dfrac{1}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)
=\(\dfrac{5}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)
=\(\dfrac{7}{6}+\dfrac{1}{81}\)
=\(\dfrac{191}{162}\)
\(2+\dfrac{1}{2}-\dfrac{4}{3}+\left(-\dfrac{1}{3}\right)^4\)
\(=\dfrac{5}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)
\(=\dfrac{5}{2}-\dfrac{108}{81}+\dfrac{1}{81}\)
\(=\dfrac{5}{2}-\dfrac{107}{81}\)
\(=\dfrac{405}{162}-\dfrac{214}{162}=\dfrac{191}{162}\)