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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 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)
= 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
4x25x0,25x1/5x1/2x2
=4x25x0,25x0,2x0,5x2
= (4x25)x(0,5x1)x(0,25x0,2)
= 100 x 1 x 0,05
= 100x0,05
= 5
\(A=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{625}+\dfrac{1}{78125}\)
\(=1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^7}\)
\(5A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}\)
\(\Leftrightarrow5A-A=5+1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^6}-1-\dfrac{1}{5}-\dfrac{1}{5^2}-\dfrac{1}{5^3}-...-\dfrac{1}{5^7}\)
\(\Leftrightarrow4A=5-\dfrac{1}{5^7}\Leftrightarrow A=\dfrac{5-\dfrac{1}{5^7}}{4}=\dfrac{\dfrac{390624}{78125}}{4}=\dfrac{390624}{312500}=\dfrac{97656}{78125}\)