1 - {1 :[23 + 1 - (-1/2)2 ]}
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Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
1. Tính:
a) 23 + 24 - ( 65 x 2 - 4 ) x 4 + 66 + 44 - ( 54 : 9 x 9 : 9 + 45 ) : 1 + 1 - 1 x ( 23 + 23 - 23 x 2 +1 ) = 51
b) 4444444444444444444444444444444444444444444444444444444444444444444 : 1
= 4444444444444444444444444444444444444444444444444444444444444444444
c) 30 = 0
d) 1/1 = 1
e)
a) - 530
b) 44444444444444444444444444444444444444444444444444444444444444444444
c) 0
d) 1
e) Em là em
a) Ta có: \(A=\sqrt{23+6\sqrt{10}}-\sqrt{23-6\sqrt{10}}\)
\(=\sqrt{18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5}-\sqrt{18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{5}\right)^2}\)
\(=3\sqrt{2}+\sqrt{5}-3\sqrt{2}+\sqrt{5}\)
\(=2\sqrt{5}\)
b) Ta có: \(B=\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+1\right)\left(\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-1\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\)
=2-1=2
\(=\dfrac{1}{2}-\left(\dfrac{1}{3\cdot7}+\dfrac{1}{7\cdot11}+...+\dfrac{1}{23\cdot27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{9-1}{27}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{8}{27}=\dfrac{1}{2}-\dfrac{2}{27}=\dfrac{27-4}{54}=\dfrac{23}{54}\)
\(\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{23}}\times\dfrac{\dfrac{1}{3}-0,25-0,2}{1\dfrac{1}{6}-0,875-0,7}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{23}-\dfrac{1}{23}}\times\dfrac{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}-\dfrac{7}{10}}\)
\(=\dfrac{\dfrac{1}{3} +\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}\times2+\dfrac{1}{7}\times2-\dfrac{1}{23}\times2}\times\dfrac{\dfrac{2}{6}-\dfrac{2}{8}-\dfrac{2}{10}}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{2\times\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}\right)}\times\dfrac{2\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{1}{2}\times\dfrac{2}{7}\)
\(=\dfrac{1}{7}\)
Bài 1:
a, Ta có: \(\dfrac{27}{23}+\dfrac{5}{21}-\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}\)
= \(\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
=\(\dfrac{23}{23}+\dfrac{21}{21}+\dfrac{1}{2}\) = \(1+1+\dfrac{1}{2}\)
= \(2\dfrac{1}{2}\)
1. \(S=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(S=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{10000}\right)\)
\(S=\frac{3}{4}.\frac{8}{9}...\frac{9999}{10000}\)
\(S=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(S=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(S=\frac{1}{100}.\frac{101}{2}\)
\(S=\frac{101}{200}\)
2.
Vì 3x - 5y \(⋮\)23
\(\Rightarrow\)6 . ( 3x - 5y ) \(⋮\)23
Ta có : 6 . ( 3x - 5y ) + ( 5x - 16y )
\(\Leftrightarrow\)( 18x - 30y ) + ( 5x - 16y )
\(\Leftrightarrow\)23x - 46y
\(\Leftrightarrow\)23 . ( x - 2y ) \(⋮\)23
Vì 18x - 30y \(⋮\)23 mà ( 5 ; 23 ) = 1
\(\Rightarrow\)5x - 16y \(⋮\)23
1- { 1 : [8 + 1 - 1/4]}
= 1 - {1 : 35/4}
= 1 - 4/35
= 31/35
1-{1:[23+1-(-1/2)2]}
=1-{1:[8+1-1/4]}
=1-{1:[9-1/4]}
=1-{1:35/4}
=1-4/35
=31/35