CMR
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\frac{5}{3^5}+.....+\frac{102}{3^{102}}<\frac{3}{4}\)
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CMR \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\frac{5}{3^5}+....+\frac{102}{3^{102}}
mjk ko bik làm nhưng có nhiều người nói bạn sai
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26 tháng 7 2015 lúc 13:03
CMR: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}+\frac{102}{3^{102}}<\frac{3}{4}\)
mik biet lam nhung ngại viet lam ban thong cam nha
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CMR \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{102}{3^{102}}\)
Giúp mình đi mà!! Cần Gấp lắm rồi
Tìm x, biết:
\(a)\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(b)\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
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a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
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a) \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Leftrightarrow\frac{x+1}{65}+\frac{x+2}{64}+2=\frac{x+3}{63}+\frac{x+4}{62}+2\)
\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+2}{64}+1\right)=\left(\frac{x+3}{63}+1\right)+\left(\frac{x+4}{62}+1\right)\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}-\frac{x+66}{63}-\frac{x+66}{62}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}-\frac{1}{62}\right)=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
CÂu b) làm tương tự:
- Trừ 3 cho hai vế ( câu a) mk cộng 2 cho hai vế)
- Tách -3 = -1-1-1 rồi kết hợp với mỗi hạng tử
CỐ LÊN NHÉ
NẾU bạn KHÔNG HIỂU thì câu b) mik sẽ làm kĩ càng và rõ ràng hơn cho bạn hiểu
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CMR: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+ \frac{1}{102}\right)=\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
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a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
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a) \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(100=\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(100=1+1+1+...+1\)
\(\Rightarrow100=100\)
b) \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
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24 tháng 2 2017 lúc 16:22
dạng 1 : so sánh a) P frac{1}{1^2}+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{2013^2}+frac{1}{2014^2}và Q 1frac{3}{4}dạng 2 : toán chứng minh 1. cho S frac{1}{101}+frac{1}{102}+...+frac{1}{130}chứng minh rằng : frac{1}{4} S frac{91}{330}2. cho S frac{5}{20}+frac{5}{21}+frac{5}{22}+...+frac{5}{49}. CMR : 3 S 83. CMR : 1+frac{1}{2}+frac{1}{3}+...+frac{1}{2^{1999}}1000
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dạng 1 : so sánh
a) P = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}\)và Q = \(1\frac{3}{4}\)
dạng 2 : toán chứng minh
1. cho S = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\)chứng minh rằng : \(\frac{1}{4}< S< \frac{91}{330}\)
2. cho S = \(\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\). CMR : 3 < S < 8
3. CMR : \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
2.a) Vào question 126036
b) Vào question 68660
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Bài 1 : Tính C frac{1}{2!}+frac{2}{3!}+frac{3}{4!}+...+frac{n-1}{n!}Bài 2 : CMR Dfrac{2!}{3!}+frac{2!}{4!}+frac{2!}{5!}+...+frac{2!}{n!} 1Bài 3: Cho biểu thức P1-frac{1}{2}+frac{1}{3}-frac{1}{4}+...+frac{1}{199}-frac{1}{200}a) CMR : P frac{1}{101}+frac{1}{102}+...+frac{1}{200}b) Giải bài toán trên trog trường hợp tổng quát Bài 4 : CMR: forall nin Zleft(nne0;nne1right) thì Q frac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{nleft(n+1right)} không phải là số nguyên .Bài 5 : CMR : Sfrac{1}{2^2}+fr...
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Bài 1 : Tính C= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n-1}{n!}\)
Bài 2 : CMR D=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}< 1\)
Bài 3: Cho biểu thức P=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
a) CMR : P= \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
b) Giải bài toán trên trog trường hợp tổng quát
Bài 4 : CMR: \(\forall n\in Z\left(n\ne0;n\ne1\right)\) thì Q= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\) không phải là số nguyên .
Bài 5 : CMR : S=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{200^2}< \frac{1}{2}\)
1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)
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