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\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
Ta có :
\(\frac{1}{10}< 1\)
\(\frac{1}{15}< 1\)
\(\frac{1}{21}< 1\)
........................
\(\frac{1}{120}< 1\)
\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)
\(\Rightarrow A< 1\)( đpcm)
Ta có : A = \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)
= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
= \(2\times\frac{3}{16}\)
= \(\frac{3}{8}\)< 1
=> A < 1
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)
\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)
c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)
do đó : A . A < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)
Tất cả dùng nhân chéo
a.3x=2.54=>x=36
b.15x=60=>x=4
c.2/3=4/6=>x=5
d.x=5;4(câu này ss từng phần)
\(a,\frac{2}{3}=\frac{x}{54}\)
\(\Rightarrow2.54=3x\)
\(\Rightarrow3x=108\)
\(\Rightarrow x=108:3=36\)
\(b,\frac{10}{x}=\frac{15}{6}\)
\(\Rightarrow10.6=15x\)
\(\Rightarrow15x=60\)
\(\Rightarrow x=60:15=4\)
\(c,\frac{2}{3}< \frac{x}{6}< 1\)
\(\Rightarrow\frac{4}{6}< \frac{x}{6}< \frac{6}{6}\)
\(\Rightarrow4< x< 6\)
\(\Rightarrow x=5\)
\(d,1< \frac{6}{x}< 2\)
\(\Rightarrow\frac{6}{6}< \frac{6}{x}< \frac{6}{3}\)
\(\Rightarrow6< x< 3\)
\(\Rightarrow x=5;4\)
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)
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