So sánh A với 1.
Biết: \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{8}{9!}+\frac{9}{10!}\)
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18 tháng 3 2018 lúc 14:40
Cho \(M=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}\)
So sánh M với 1
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
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Ta có:
1/2=0,5
2/3>0,6
<=>1/2+2/3>1,1>1
<=>1/2+2/3+3/4+...+9/10>1
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Vì 1 = \(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
\(\Rightarrow\)M > 1 vì \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)
\(\Rightarrow M>1\)
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So sánh \(A\)với\(13\),biết rằng:
\(A=\frac{13}{15}+\frac{7}{5}+\frac{3}{4}+\frac{1}{5}+\frac{1}{7}+\frac{19}{20}+\frac{5}{4}+\frac{1}{3}+\frac{1}{6}+\frac{1}{13}+\frac{17}{23}+\frac{9}{8}+\frac{2}{5}+\frac{1}{7}+\frac{1}{25}+\frac{3}{2}+\frac{1}{8}+\frac{1}{19}+\frac{1}{9}+\frac{1}{97}\)
Tìm A:B, biết:
A=\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\)
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
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\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
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Cho \(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
So sánh S với 10
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
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A left(frac{1}{2}+frac{1}{4}+frac{1}{8}-1right)timesleft(1-frac{8}{1}-frac{4}{1}-frac{2}{1}right)B frac{frac{3}{1}-frac{6}{3}-frac{9}{6}-frac{369}{1}}{frac{1}{3}+frac{3}{6}+frac{6}{9}-frac{1}{963}}C frac{1}{1}-frac{1}{2}+frac{3}{1}-frac{1}{4}+frac{5}{1}-frac{1}{6}+frac{7}{1}-frac{1}{8}+frac{9}{1}-frac{1}{10}so sánh các số trên ( A , B , C )
Đọc tiếp
A = \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-1\right)\times\left(1-\frac{8}{1}-\frac{4}{1}-\frac{2}{1}\right)\)
B = \(\frac{\frac{3}{1}-\frac{6}{3}-\frac{9}{6}-\frac{369}{1}}{\frac{1}{3}+\frac{3}{6}+\frac{6}{9}-\frac{1}{963}}\)
C = \(\frac{1}{1}-\frac{1}{2}+\frac{3}{1}-\frac{1}{4}+\frac{5}{1}-\frac{1}{6}+\frac{7}{1}-\frac{1}{8}+\frac{9}{1}-\frac{1}{10}\)
so sánh các số trên ( A , B , C )
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
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A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
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Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
bài 1 So sánha)Afrac{3}{8^3}+frac{7}{8^4} ; Bfrac{7}{8^3}+frac{3}{8^4}b)Afrac{10^{1992}+1}{10^{1991}+1};Bfrac{10^{1993}+1}{10^{1992}+1}c)Afrac{10^7+5}{10^4-8};Bfrac{10^8+6}{10^8-7}d)Afrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};Bfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}e)Afrac{2011}{2012}+frac{2012}{2013};Bfrac{2011+2012}{2012+2013}
Đọc tiếp
bài 1 So sánh
a)\(A=\frac{3}{8^3}+\frac{7}{8^4}\) ; \(B=\frac{7}{8^3}+\frac{3}{8^4}\)
b)\(A=\frac{10^{1992}+1}{10^{1991}+1};B=\frac{10^{1993}+1}{10^{1992}+1}\)
c)\(A=\frac{10^7+5}{10^4-8};B=\frac{10^8+6}{10^8-7}\)
d)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
e)\(A=\frac{2011}{2012}+\frac{2012}{2013};B=\frac{2011+2012}{2012+2013}\)
Bài 1:Tìm x biết Bài 2:So sánh
a, x+frac{1}{2}frac{3}{8}.frac{4}{5} a, Afrac{10^{10}-1}{10^{11}-1}vaBfrac{10^9-1}{10^{10}-1}
b, frac{5}{16}:x-frac{1}{4}frac{5}{8} b, B frac{10^{10}}{10^{10}+1}vaBfrac{10^{10}+1}{10^{10}+2}
c, frac{-1}{4}.x+frac{3}{7}.x2
d, frac{22}{9}-left(x+frac{1}{2}right)^2frac{7}{3}
e, left|frac{1}{4}-xright|+5frac{1}{8}6frac{1}{8}
Đọc tiếp
Bài 1:Tìm x biết Bài 2:So sánh
a, \(x+\frac{1}{2}=\frac{3}{8}.\frac{4}{5}\) a, \(A=\frac{10^{10}-1}{10^{11}-1}vaB=\frac{10^9-1}{10^{10}-1}\)
b, \(\frac{5}{16}:x-\frac{1}{4}=\frac{5}{8}\) b, B =\(\frac{10^{10}}{10^{10}+1}vaB=\frac{10^{10}+1}{10^{10}+2}\)
c, \(\frac{-1}{4}.x+\frac{3}{7}.x=2\)
d, \(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
e, \(\left|\frac{1}{4}-x\right|+5\frac{1}{8}=6\frac{1}{8}\)
a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
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