Ch/m 1/2^2 +1/3^2 +1/4^2+...+1/100^2 <1
(BẠN NÀO TỐT GIÚP MÌN)
Những câu hỏi liên quan
M=1 + 1/2 (1+2) + 1/3 (1+2+3) +1/4 (1+2+3+4) +...+ 1/100. (1+2+3+...+100) = ?
Tính:
M=(1-1/2^2).(1-1/3^2).(1-1/4^2)...(1-1/49^2).(1-1/50^2)
N=(3/2-2/2^2).(4/3-2/3^2).(5/4-2/4^2)...(100/99-2/99^2).(101/100-2/100^2)
??????????????????????????????????????????
c/m
a/
1/2!+2/3!+3/4!+...+99/100!<1
b/
1*2-1/2!+2*3-1/3!+3*4-1/4!+...+99*100-1/100!<2
1
(-1).(-1)2.(-1)3.(-1)4......(-1)100
25. (75-45) - 75.(45 - 25) làm 2 cách
2 tìm x
-2 (2x -3 ) +5 (-3 -x = -3 . ( x -0)
1)
25.(75-45)-75.(45-25)
C1: =25.30-75.20
=750-1500
=-750
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(-1).(-1)\(^2\).(-1)\(^3\).....(-1)\(^{100}\)
\(\Rightarrow\)(-1).1.(-1).1.....(-1).1
có tất cả 50 số -1
có tất cả 50 số 1
\(\Rightarrow\) \([\)(-1).50\(]\).\([\)1.50\(]\)
=-50.50=0
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Rút gọn
A= 2^100+2^99+2^98.....+2+1
B=3^100+3^99+3^98....+3+1
C=4^100+4^99+....+4+1
D=2^100- 2^99+....+2^2 - 2 + 1
E=3^100 - 3^99 + 3^98....- 3 +1
Thu gọn
M= 2 + 2^2 + 2^3 ....+ 2^100
Cho A =2+2^2+2^3+....2^100. Tìm số tự nhiên x sao cho A + 1 = 2x
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
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giá trị của tích (1-1/2^2)(1-1/3^2)(1-1/4^2).....(1-1/99^)(1-1/100^2)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{98}{99}.\dfrac{100}{99}.\dfrac{99}{100}.\dfrac{101}{100}\)
\(=\dfrac{1.2...98.99}{2.3...99.100}.\dfrac{3.4...100.101}{2.3...99.100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
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Chứng minh rằng: \(\dfrac{1}{3^2}^{ }+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Làm nhanh và hay mình tích cho.
A = \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\)
A < \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
A < \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A < \(\dfrac{1}{2}-\dfrac{1}{100}\)
⇒ A < \(\dfrac{1}{2}\)
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1. Tìm x
a)/ x-3/2 /+ / x+1 /+/ x-2 /= 4x
b)/ 2x-1 /+/ x+3 /=5
c)/ 3x-1 /+/ 3x+1 /=/ x-2 /
2. Tìm M
M=1+ 3/2^3+ 4/2^4+...+100/2^100
1.a) có: \(|x-\frac{3}{2}|,|x+1|,\left|x-2\right|\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(x\ge0\Rightarrow x-\frac{3}{2}\ge\frac{-3}{2}\Rightarrow\left|x-\frac{3}{2}\right|\ge\left|\frac{-3}{2}\right|=\frac{3}{2}\Rightarrow\left|x-\frac{3}{2}\right|=x-\frac{3}{2}\)
cmtt: \(|x-2|=x-2\)
\(\Rightarrow3x-\frac{3}{2}+1-2=4x\)
\(\Rightarrow3x-\frac{5}{2}=4x\)
\(\Rightarrow x=\frac{-5}{2}\left(ko,t/m\right)\)
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\(M=\frac{2^3-1}{2^3+1}.\frac{3^3-1}{3^3+1}.\frac{4^3-1}{4^3+1}....\frac{100^3-1}{100^3+1}\)
CHỨNG MINH M> 2/3
Ta có : \(\frac{a^3-1}{\left(a+1\right)^3+1}=\frac{\left(a-1\right)\left(a^2+a+1\right)}{\left(a+1+1\right)\left(\left(a+1\right)^2-\left(a+1\right)+1\right)}=\frac{a-1}{a+2}\)
\(M=\frac{100^3-1}{2^3+1}.\frac{2^3-1}{3^3+1}.\frac{3^3-1}{4^3+1}...\frac{99^3-1}{100^3+1}\)
\(M=\frac{999999}{9}.\frac{1}{4}.\frac{2}{5}.\frac{3}{6}...\frac{98}{101}=\frac{999999.1.2.3}{9.99.100.101}\)
\(M=\frac{10101.2}{3.100.101}=\frac{20202}{30300}>\frac{20200}{30300}=\frac{2}{3}\)