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A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
M=100
Xét tử N
92-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=(1+1+1+...+1)-(1/9)-(2/10)-(3/11)- ... -(90/98)-(91/99)-(92/100)
=1-(1/9)+1-(2/10)+1-(3/11)+......+1-(90/98)+1-(91/99)+1-(92/100)
=(8/9)+(8/10)+(8/11)+ ...+ (8/98)+(8/99)+(8/100)
=8.[(1/9)+(1/10)+(1/11)+...+(1/98)+(1/99)+(1/100)]
=40[(1/45)+(1/50)+(1/55)+...+(1/495)+(1/500)]
=>N=40
=>M/N=5/2
Cho : B=\(\frac{4}{3}\)+\(\frac{10}{9}\)+\(\frac{28}{27}\)+...+\(\frac{3^{98}+1}{3^{98}}\)CM : B<100
\(B=\frac{3^n+1}{3^n}=1+\frac{1}{3^n}=C+D\)
B có 98 số hạng => C=98
\(D=\frac{1}{3}+\frac{..1}{3^{97}}+\frac{1}{3^{98}}\)
3.D=1+1/3+....+1/3^97
tRỪ CHO NHAU
2D=1-1/3^98
\(C=\frac{1}{2}-\frac{1}{2.3^{98}}< \frac{1}{2}\)
\(B=98+\frac{1}{2}-\frac{1}{2.3^{98}}< 99< 100\) có lẽ đề lấy 100 co chẵn. hay cộng nhầm ai tets hộ cái
\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(=\frac{3+1}{3}+\frac{3^2+1}{3^2}+\frac{3^3+1}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)
\(=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)
\(=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
\(\text{Đặt }A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\text{rút gon cái A thì dc: }A=\frac{1}{3^{98}}-1\Rightarrow B=98+\frac{1}{3^{98}}-1=97+\frac{1}{3^{98}}\)
\(
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}\)
\(B=\frac{1}{100}\)
mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)
Đặt \(A=\frac{\frac{2000}{11}+\frac{2000}{12}+...+\frac{2000}{100}}{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}\)
\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{100.\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)}\)
\(\Rightarrow A=\frac{20.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}}\)
\(\Rightarrow A=\frac{\frac{20}{11}+\frac{20}{12}+..+\frac{20}{100}}{\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}+\frac{1}{100}}\)
A=\(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)
=> A>\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\) = 1+1+..+1 =98
A=\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\) +\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)> 1+1+..+1 = 98
Đặt B = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
=>2B = 1-\(\frac{1}{3^{98}}\) <1
=> B<1
=>A<99
=>98<A<99