CMR:\(\dfrac{7}{12}< \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{5}{6}\)
giúp mk nhé
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a=(\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)):(\(\dfrac{-6}{51}-\dfrac{6}{52}-\dfrac{6}{53}-...-\dfrac{6}{100}\))
giúp mik giải nhé
cảm ơn !
a) dfrac{2}{1^2}.dfrac{6}{2^2}.dfrac{12}{3^2}.dfrac{20}{4^2}.dfrac{30}{5^2}.....dfrac{110}{10^2}.x-20
b) left(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{2013}right).x+2013dfrac{2014}{1}+dfrac{2015}{2}+...+dfrac{4025}{2012}+dfrac{4026}{2013}
c) left(dfrac{1}{1.2}+dfrac{1}{3.4}+...+dfrac{1}{99.100}right).xdfrac{2012}{51}+dfrac{2012}{52}+...+dfrac{2012}{99}+dfrac{2012}{100}
Đọc tiếp
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
Tham khảo: (mk chx chắc lắm đâu nha)
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Bài 1: Cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
a) Chứng minh: A=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
b) Chứng minh: A<\(\dfrac{5}{6}\)
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
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\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}:\dfrac{1}{1-2}+\dfrac{1}{2-3}+...+\dfrac{1}{99-100}\)
9 - 3 x ( X - 9 ) 64 + 6 x ( X + 1 ) 70dfrac{X}{13}+dfrac{15}{26}dfrac{46}{52}dfrac{11}{14}-dfrac{3}{X}dfrac{5}{14}5 x ( 3 + 7 x X ) 40X x 6 + 12 : 3 120X x 3,7 + X x 6,3 120( 15 x 24 - X ) : 0,25 100 : dfrac{1}{4}71 + 65 x 4 dfrac{X+140}{X}+ 260( X +1 ) + ( X + 4 ) + ( x + 7 ) + ...... + (X + 28 ) 155đây là bài tìm X
Đọc tiếp
9 - 3 x ( X - 9 ) = 6
4 + 6 x ( X + 1 ) 70
\(\dfrac{X}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{11}{14}-\dfrac{3}{X}=\dfrac{5}{14}\)
5 x ( 3 + 7 x X ) = 40
X x 6 + 12 : 3 = 120
X x 3,7 + X x 6,3 = 120
( 15 x 24 - X ) : 0,25 = 100 : \(\dfrac{1}{4}\)
71 + 65 x 4 = \(\dfrac{X+140}{X}\)+ 260
( X +1 ) + ( X + 4 ) + ( x + 7 ) + ...... + (X + 28 ) = 155
đây là bài tìm X
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
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Chứng tỏ
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Đặt A= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)
= \(\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\right)\)
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rút gọn
a.\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b.\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
lm mhanh giúp mk nhé!mk đang cần gấp!
`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`
`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`
`=sqrt{10}-(2+sqrt{10})`
`=-2`
`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`
`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`
`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`
`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`
`=(5sqrt5-5sqrt3-4)/2`
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\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}:\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)