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( 1 + 1/2 ) . ( 1+ 1/3) . ( 1+ 1/4 ) . ( 1+ 1/5 )
=3/2 . 4/3.5/4.6/5
= 3.4.5.6/2.3.4.5
=6/2 = 3
= 1 x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{5}\) )
= 1 x \(\dfrac{77}{60}\)
= \(\dfrac{77}{60}\)
Lời giải:
Gọi tích trên là $A$. Ta có:
$A=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{5}{6}$
$=\frac{1\times 2\times 3\times 4\times 5}{2\times 3\times 4\times 5\times 6}=\frac{1}{6}$
Bài 3
a,26/100+0,009+41/100+0,24
0,26+0,09+0,41+0,24
(0,26+0,24)+(0,09+0,41)
0,5+0,5
=1
b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4
(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)
30+1+1+1+1
=34
Bài 4,5 khó quá mik ko bít lamf^^))
Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1
\(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)
b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)
Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)
Giải:
\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
15/7 : 1/2 + 3/14 : 1/2
= 15/7 . 2 + 3/14 . 2
= (15 + 3/14) . 2
= 213/14 . 2
= 213/7
\(\dfrac{15}{7}\) : \(\dfrac{1}{2}\) + \(\dfrac{3}{14}\) : \(\dfrac{1}{2}\)
= (\(\dfrac{15}{7}\) + \(\dfrac{3}{14}\)) : \(\dfrac{1}{2}\)
= (\(\dfrac{15\times2}{7\times2}\) + \(\dfrac{3}{14}\)) : \(\dfrac{1}{2}\)
= (\(\dfrac{30}{14}\) + \(\dfrac{3}{14}\)) \(\times\) 2
= \(\dfrac{33}{14}\) \(\times\) 2
= \(\dfrac{33}{7}\)
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)
= 4/3*9/8*16/15*25/24*36/35
=2*2/1*3 * 3*3/2*4 *4*4/3*5 *5*5/4*6 * 6*6/5*7
= (2*3*4*5*6 / 1*2*3*4*5) * ( 2*3*4*5*6 / 3*4*5*6*7)
=6/1* 2/7
= 12/7
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}\)
\(=\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}\)
\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}\)
\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{21}=\dfrac{38}{21}\)