\(Chứngtỏrằng:\frac{1}{^22}+\frac{1}{^23}+\frac{1}{^24}+\frac{1}{^25}+\frac{1}{^26}+\frac{1}{^27}+\frac{1}{^28}< 1\)
chứng tỏ rằng 1 phần 2 mũ 2+1 phần ba mũ 2...........
Những câu hỏi liên quan
\(S=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}\)\(\frac{1}{30}\)
Hãy so sánh S với \(\frac{1}{3}\)
ta có 1/3=10/30
1/21+1/22+...+1/30 có 10 p/số
mà 1/21>1/30
1/22>1/30
....
1/29>1/30
1/30=1/30
=>1/21+..1/30>1/30+....1/30 có 10 phân số
=>1/21+...1/30>1/3
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Ta có: \(\frac{1}{21}< \frac{1}{30}\)
\(\frac{1}{22}< \frac{1}{30}\)
......
\(\frac{1}{29}< \frac{1}{30}\)
\(\Rightarrow S< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 p/s)
\(\Rightarrow S< \frac{1}{30}.10=\frac{10}{30}=\frac{1}{3}\)
Vậy S < 1/3
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ta co 1/21+1/22+1/23>3/30
1/24+1/25+1/26>3/30
1/27+1/28+1/29>3/30
==>S>3/30+3/30+3/30+1/30
S>10/30 hay S>1/3
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a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27}Leftrightarrowfrac{x-23}{24}+frac{x-23}{25}-frac{x-23}{26}-frac{x-23}{27}0Leftrightarrowleft(x-23right)left(frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}right)0Leftrightarrow x-230( vì frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}ne0)Leftrightarrow x23Vậy nghiệm của pt x23
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a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\)( vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\))
\(\Leftrightarrow x=23\)
Vậy nghiệm của pt x=23
So sánh\(\frac{1}{2}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}+\frac{1}{26}\)và 1
ChoA=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}....+\frac{1}{49}-\frac{1}{50}\)
B=\(\frac{1}{25}+\frac{1}{26}+\frac{1}{27}.....+\frac{1}{50}\)
Chứng tỏ rằng A=B
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\Rightarrow A=B\text{(đpcm)}\)
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bài này chắc mình không làm được rồi, xin lỗi
hihi
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cho tổng:N frac{1}{21}+ frac{1}{22}+ frac{1}{23}+ frac{1}{24}+ frac{1}{25}+ frac{1}{26}+ frac{1}{27}+ frac{1}{28}+ frac{1}{29}+ frac{1}{30} . Chứng tỏ N frac{1}{3}.hơi xoắn não mọi người ạ!!!
Đọc tiếp
cho tổng:
N = \(\frac{1}{21}\)+ \(\frac{1}{22}\)+ \(\frac{1}{23}\)+ \(\frac{1}{24}\)+ \(\frac{1}{25}\)+ \(\frac{1}{26}\)+ \(\frac{1}{27}\)+ \(\frac{1}{28}\)+ \(\frac{1}{29}\)+ \(\frac{1}{30}\) . Chứng tỏ N > \(\frac{1}{3}\).
hơi xoắn não mọi người ạ!!!
Câu 1: Chứng tỏ rằng:
\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{26}+\frac{1}{27}>\frac{8}{27}\)
Câu 2: Tính:
\(D=\frac{2^{10}.6^{15}+3^{14}.15.4^{13}}{2^{18}.18^7.3^3+3^{15}.2^{25}}\)
#)Giải :
Câu 1 :
Đặt \(A=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{27}\)
\(\Rightarrow A>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}\)( 8 số hạng )
\(\Rightarrow A>\frac{8}{27}=\frac{8}{27}\)
\(\Rightarrow A>\frac{8}{27}\)
#~Will~be~Pens~#
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Câu 1:(trội)
Ta có:\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}=\frac{8}{27}\left(đpcm\right)\)
Câu 2:\(D=\frac{2^{25}.3^{15}+3^{15}.5.2^{26}}{2^{25}.3^{17}+3^{15}.2^{25}}=\frac{2^{25}3^{15}\left(1+5.2\right)}{2^{25}3^{15}\left(3^2+1\right)}=\frac{11}{10}\)
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Chứng minh rằng : \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Mk làm câu a thôi nhé :)
Vì \(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(=>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(< \)\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)(1)
Vì \(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
...
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(=>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}\)(2)
Từ (1) và (2) => ĐPCM
Chứng minh rằng
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
Ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
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Chứng minh rằng :
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{49}+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\) (đpcm)
*đpcm = điều phải chứng minh
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