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Sử dụng ''2'' ta có;
\(\frac{n}{n+1}.\frac{n+3}{n+1}=\frac{n^2+2n+n}{n^2+2n+1}\ge1.\)
Suy ra
\(\frac{n}{n+1}\) lớn hơn \(\frac{n+1}{n+3}.\)
P/s; Sao ko ai giúp vậy huhu ToT
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 10^30 và 2^100
\(10^{30}=10^{3.10}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=2^{10.10}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)
b) 5^36 và 11^24
\(5^{36}=5^{3.12}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=11^{2.12}=\left(11^2\right)^{12}=121^{12}\)
Vì ........................ nên .................................
c) 9^12 và 27^7
\(9^{12}=\left(3^2\right)^{12}=3^{2.12}=3^{24}\)
\(27^7=\left(3^3\right)^7=3^{3.7}=3^{21}\)
Vì ............................ nên .................................
a)10^30và2^100
10^30=(10^3)^10=1000^10
2^100=(2^10)^10=1024^10
Vì 1024>1000 nên 2^100>10^30
b)5^36 và 11^24
5^36=(5^3)^12=125^12
11^24=(11^2)^12=121^12
Vì 125>121nên 5^36>11^24
c)9^12 và 27^7
9^12=(3^2)^12=3^24
27^7=(3^3)^12=3^36
Vì 24>21 nên 9^12>27^7
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\(A=\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =\dfrac{2}{4}.\left(\dfrac{1}{3}-\dfrac{1}{n+4}\right)\\ =\dfrac{1}{2}.\dfrac{n+1}{3\left(n+4\right)}=\dfrac{n+1}{6\left(n+4\right)}\\ =\dfrac{n+4-3}{6\left(n+4\right)}=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}< \dfrac{1}{6}.\)
Giải:
A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)
A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)
A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)
A=1/2.(1/3-1/n+4)
A=1/6-1/2.(n+4)
⇒A>1/6
Chúc bạn học tốt!
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1) ta có:\(2^{150}\)= (2^3)^50=8^50
\(3^{100}\)= (3^2)^50 = 9^50
vì 8^50 < 9^50 => \(2^{150}\)<\(3^{100}\)
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\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{n}-\dfrac{1}{n+k}=\dfrac{n+k}{n\left(n+k\right)}-\dfrac{n}{n\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{k}{n\left(n+k\right)}\)
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)(đpcm)
Sử dụ ''2'' ta có
\(\dfrac{n}{n+1}.\dfrac{n+1}{n+3}=\dfrac{n^2+2n+n}{n^2+2n+1}\ge1.\)
Suy ra
\(\dfrac{n}{n+1}\) lớn hơn \(\dfrac{n+1}{n+3}\) \(\in N\)
Mk mới học mong các bạn giúp đỡ