Cho P= \(\frac{1}{3}+\frac{1}{16}+\frac{1}{19}+\frac{1}{21}+\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\)
So sánh P với \(\frac{3}{5}\)
Những câu hỏi liên quan
Cho B=\(\frac{1}{3}+\frac{1}{16}+\frac{1}{19}+\frac{1}{21}+\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\)
So sánh B với \(\frac{3}{5}\)
12 tháng 7 2018 lúc 16:06
Bài 3
Cho B =\(\frac{1}{3}+\frac{1}{16}+\frac{1}{19}+\frac{1}{21}+\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\)
So sánh B=\(\frac{3}{5}\)
Ta có : 1/61 +1/72 +1/83 +1/94 < 1/60 x4 =1/15
1/16 +1/19 +1/21 < 1/15 x3 =3/15
1/3 =5/15
Do đó: B < 1/15 + 3/15 +5/15 = 9/15= 3/5
Vậy B< 3/5
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\(\text{Bài giải}\)
\(\text{Ta có : }\)
\(B=\frac{1}{3}+\frac{1}{16}+\frac{1}{19}+\frac{1}{21}+\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\)
B = 0,333333333 + 0,0625 + 0,0526315789 + 0,0476190476 + 0,0163934426 + 0,0138888889 + 0,0120481928 + 0,0106382979
B = 0,549052782
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Bài 1 : cho 2 biểu thứcAleft(1-frac{1}{2}right)left(1-frac{1}{3}right)...left(1-frac{1}{19}right)left(1-frac{1}{20}right)Bleft(1-frac{1}{4}right)left(1-frac{1}{9}right)left(1-frac{1}{16}right)...left(1-frac{1}{81}right)left(1-frac{1}{100}right)So sánh A với frac{1}{21}So sánh B với frac{11}{21}
Đọc tiếp
Bài 1 : cho 2 biểu thức
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
So sánh A với \(\frac{1}{21}\)
So sánh B với \(\frac{11}{21}\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
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Thực hiện so sánh:\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\)\(+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}\)\(+\frac{1}{23}\)với \(\frac{5}{6}\)
Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
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Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
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1. tính A frac{3}{2^2}.frac{8}{3^2}.frac{15}{4^2}...frac{899}{30^2}2. tính B frac{1}{4}.frac{2}{6}.frac{3}{8}.frac{4}{10}...frac{30}{62}.frac{31}{64}3. So sánh C left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)...left(1-frac{1}{20}right)với frac{1}{21}4. So sánh D left(1-frac{1}{4}right).left(1-frac{1}{9}right).left(1-frac{1}{16}right)...left(1-frac{1}{100}right)với frac{11}{19}
Đọc tiếp
1. tính A= \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
2. tính B= \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
3. So sánh C= \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)với \(\frac{1}{21}\)
4. So sánh D= \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\)với \(\frac{11}{19}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
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So sánh \(A\)với\(13\),biết rằng:
\(A=\frac{13}{15}+\frac{7}{5}+\frac{3}{4}+\frac{1}{5}+\frac{1}{7}+\frac{19}{20}+\frac{5}{4}+\frac{1}{3}+\frac{1}{6}+\frac{1}{13}+\frac{17}{23}+\frac{9}{8}+\frac{2}{5}+\frac{1}{7}+\frac{1}{25}+\frac{3}{2}+\frac{1}{8}+\frac{1}{19}+\frac{1}{9}+\frac{1}{97}\)
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
b)\(\frac{3}{7}.19.\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
c)\(15\frac{1}{4}:(\frac{-5}{7})-25\frac{1}{4}:(\frac{-5}{7})\)
a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=1\frac{4}{23}-\frac{4}{23}+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
b) \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)=-6\)
c) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
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a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2+0,5\)
\(=2,5.\)
b) \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6.\)
c) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(\frac{61}{4}-\frac{101}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14.\)
Chúc bạn học tốt!
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1. so sánh
A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+với1\)
B=\(1-\left(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)với\frac{1}{2}\)
C=\(1-\left(\frac{1}{5}+\frac{1}{11}+\frac{1}{10}+\frac{1}{9}+\frac{1}{59}+\frac{1}{58}+\frac{1}{57}\right)với\frac{1}{2}\)
Cho Afrac{frac{1}{2001}+frac{1}{2002}+frac{1}{2003}+...+frac{1}{4000}}{frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{3999.4000}}. Bfrac{left(17+1right).left(frac{17}{2}+1right).left(frac{17}{3}+1right).....left(frac{17}{19}+1right)}{left(1+frac{19}{17}right).left(1+frac{19}{16}right).left(1+frac{19}{15}right).....left(1+19right)}So sánh A-Bvới 0.LÀM ĐÚNG ĐƯỢC CHO TICK
Đọc tiếp
Cho \(A=\frac{\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{3999.4000}}\).
\(B=\frac{\left(17+1\right).\left(\frac{17}{2}+1\right).\left(\frac{17}{3}+1\right).....\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right).\left(1+\frac{19}{16}\right).\left(1+\frac{19}{15}\right).....\left(1+19\right)}\)
So sánh \(A-B\)với \(0\).
LÀM ĐÚNG ĐƯỢC CHO TICK
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
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xin lỗi, mk mới học lớp 5 nên ko giúp đc gì, mong cậu thông cảm, chúc cậu học giỏi nha
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