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\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2023\cdot2024}\)
\(=\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2023}-\frac{1}{2024}\)
\(=\frac11-\frac{1}{2024}=\frac{2023}{2024}\)
A=1⋅21+2⋅31+3⋅41+⋯+2023⋅20241
\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \hdots + \frac{1}{2023} - \frac{1}{2024}\)
\(= \frac{1}{1} - \frac{1}{2024} = \frac{2023}{2024}\)
A=1/(1+3)+1/(1+3+5)+1/(1+3+5+7)+...+1/(1+3+5+7+...+2017)
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
2A=2/2^2+2/3^2+2/4^2+...+2/1009^2
Ta co :(x-1)(x+1)=(x-1)x+x-1=x^2-x+x-1=x^2-1<x^2
suy ra 2A<2/(1*3)+2/(3*5)+2/(5*7)+...+2/(1008*1010)
suy ra 2A <1-1/3+1/3-1/5+1/5-1/7+...+1/1008-1/1010
suy ra 2A<1-1/1010
suy ra 2A<2009/2010<1<3/2
suy ra 2A <3/2
suy ra A <3/4 (dpcm)
nho k cho minh voi nha
a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2
1 = 1 = 1
1/2^2 < 1/1.2 = 1/1 - 1/2
1/3^2 < 1/2.3 = 1/2 - 1/3
.......................
1/100^2 < 1/99.100 = 1/99 - 1/100
Cộng vế với vế ta có:
A = 1 + 1 - 1/100
A = 2 - 1/100 < 2 (đpcm)
\(...\Leftrightarrow-\frac{1}{10}< x< \frac{3}{5}\)
\(-\frac{1}{10}< x\Rightarrow-\frac{1}{10}< \frac{10x}{10}\Rightarrow10x>1\Rightarrow x>\frac{1}{10}\) (*)
\(x< \frac{3}{5}\Rightarrow\frac{5x}{5}< \frac{3}{5}\Rightarrow5x< 3\Rightarrow x< \frac{3}{5}\)
Vậy \(\frac{1}{10}< x< \frac{3}{5}\)
Ta có: \(1+3=4=2^2=\left(\frac{1+3}{2}\right)^2\)
\(1+3+5=9=3^2=\left(\frac{1+5}{2}\right)^2\)
....
\(1+3+5+\cdots+2023=\left(\frac{2023+1}{2}\right)^2=1012^2\)
Do đó: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{1012^2}\)
Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)
...
\(\frac{1}{1012^2}<\frac{1}{1011\cdot1012}=\frac{1}{1011}-\frac{1}{1012}\)
Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{1012^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{1011}-\frac{1}{1012}\)
=>\(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{1012^2}<\frac12-\frac{1}{1012}\)
=>\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{1012^2}<\frac14+\frac12-\frac{1}{1012}\)
=>\(A<\frac34-\frac{1}{1012}\)
=>\(A<\frac34\)
A=11+3+11+3+5+11+3+5+7+...+11+3+5+7+...+2023
A=122+132+142+...+110122
Nhận xét:
132<12.3
142<13.4
...
110122<11011.1012
⇒A<122+12.3+13.4+...+11011.1012
⇒A<14+12−13+13−14+...+11011−11012
⇒A<14+12−11012
⇒A<34−11012<34
Vậy A<34(đpcm)
b)B=12−14+18−116+132−164
B=12−122+123−124+125−126
2B=2.(12−122+123−124+125−126)
2B=1−12+122−123+124−125
2B+B=(1−12+122−123+124−125)+(12−122+123−124+125−126)
3B=1−126
B=13−126.3<13
Vậy B<13(đpcm)
c)C=131+132+133+...+160
C=(131+132+133+...+145)+(146+147+148+...+160)
Nhận xét:
131+132+133+...+145>145+145+145+...+145=13
146+146+147+...+160>160+160+160+...+160=14
⇒C>13+14
⇒C>412+312
⇒C>712
Vậy C>712(đpcm)