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Cho A = 1/32 + 1/33 + 1/34 + ... + 1/39
=>3A=1/3+1/32+1/33+...+1/38
=>3A-A=1/3+1/32+1/33+...+1/38-1/32-1/33-1/34-...-1/39
=>2A=1/3-1/39
=>\(A=\frac{\frac{1}{3}-\frac{1}{3^9}}{2}\)<1
Vậy A<1
Ta có : 2x + 2x + 1 = 24
=> 2x(1 + 2) = 24
=> 2x.3 = 24
=> 2x = 8
=> 2x = 23
=> x = 3
Ta có : (x + 2)4 = (x + 2)6
=> (x + 2)4 - (x + 2)6 = 0
<=> (x + 2)4 (1 - (x + 2)2) = 0
<=> \(\orbr{\begin{cases}\left(x+2\right)^4=0\\\left(1-\left(x+2\right)^2\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\\left(x+2\right)^2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}<\frac{1}{1.1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}<2\)
Ta có: A < \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\) (1)
Lại có: \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=1+\left(1-\frac{1}{50}\right)=1+\frac{49}{50}\)
Mà 1+49/50 < 2 (2)
Từ (1) và (2) ta có: A<1+49/50<2
Vậy A<2
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{50^2-1}\)
\(=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{50}+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{50}-\frac{1}{51}\right)< \frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}\left(dpcm\right)\)
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(=3^0-3^1+3^2-3^3+...+3^{98}-3^{99}\)có 100 hạng tử
\(=\left(3^0-3^1+3^2-3^3\right)+\left(3^4-3^5+3^6-3^7\right)+...+\left(3^{96}-3^{97}+3^{98}-3^{100}\right)\) có 25 cặp
\(=-20+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)\)
\(=-20\left(1+3^4+...+3^{96}\right)⋮-20\)
ta co
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}>\frac{1}{2.2}+\frac{1}{3.3}+....+\frac{1}{10.10}\)
ma ve trai =\(1-\frac{1}{10}\)
nen ve phai <1
D=1/1.2+1/2.3+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
=1+0+0+0+...+0-1/10=1-1/10=9/10
ta có ; 1/22 +1/32+...+1/20172<1/1.2+1/2.3+1/3.4+.....+1/2016.2017=1-1/2+1/2-1/3+...+1/2016-1/2017=1+0+0+0+...+0-1/2017
=1-1/2017<1