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Ta có: A=1/11+1/12+1/13+...+1/30
=(1/11+1/12+1/13+..+1/20)+(1/21+1/22+1/23+...+1/30)
\(\Rightarrow\)A<(1/10+1/10+1/10+...+1/10)+(1/20+1/20+1/20+...1/20)
\(\Rightarrow\)A<(1/10)*10+(1/20)*10
\(\Rightarrow\)A<1+1/2
\(\Rightarrow\)A<3/2<11/6
Ta thấy : \(\frac{1}{11}>\frac{1}{100},\frac{1}{12}>\frac{1}{100},...,\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}=1\)
Do đó : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>1\)
mình nhầm câu b:
Áp dụng....
A=10^11-1/10^12-1<10^11-1+11/10^12-1+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)
=10^10+1/10^11+1=B
Vậy A<B(câu này mới đúng còn câu b mình làm chung với câu a là sai)
a) Với a<b=>a+n/b+n >a/b
Với a>b=>a+n/b+n<a/b
Với a=b=>a+n/b+n=a/b
b) Áp dụng t/c a/b<1=>a/b<a+m/b+m(a,b,m thuộc z,b khác 0)ta có:
A=(10^11)-1/(10^12)-1=(10^11)-1+11/(10^12)-1+11=(10^11)+10/(10^12)+10=10.[(10^10)+1]/10.[(10^11)+1]
=(10^10)+1/(10^11)+1=B
Vậy A=B
\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)
Học good
\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}\)
\(=-\frac{101}{200}\)
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>10.\frac{1}{20}+10.\frac{1}{30}\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)
Chúc bạn học tốt ~
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>\frac{1}{20}\times10+\frac{1}{30}\times10\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)