Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(6:3,5-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\\ =\left(\dfrac{12}{7}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\\ =\left(\dfrac{17}{7}-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)\\ =\dfrac{10}{7}:9\\ =\dfrac{10}{63}\)
\(\left(6:\dfrac{3}{5}-1\dfrac{1}{6}\times\dfrac{6}{7}\right):\left(4,2\times\dfrac{10}{11}+5\dfrac{2}{11}\right)\)
\(=\left(6\times\dfrac{5}{3}-\dfrac{7}{6}\times\dfrac{6}{7}\right):\left(\dfrac{21}{5}\times\dfrac{10}{11}+\dfrac{57}{11}\right)\)
\(=\left(10-1\right):\left(\dfrac{42}{11}+\dfrac{57}{11}\right)=9:9=1\)
Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$
$x-x\times \frac{1}{2}=\frac{3}{2}$
$x\times (1-\frac{1}{2})=\frac{3}{2}$
$x\times \frac{1}{2}=\frac{3}{2}$
$x=\frac{3}{2}: \frac{1}{2}=3$
(1-1/2)x(1-1/3)x(1-1/4)x.....x(1-1/2015)x(1-1/2016)
=> 1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9 x......... x 2014/2015 x 2015/2016
Ta rút gọn cho tử này mẫu kia còn: 1/2016.
Đáp số : 1/2016
\(=\frac{2-1}{2}\times\frac{3-1}{3}\times\frac{4-1}{4}\times...\times\frac{2015-1}{2015}\times\frac{2016-1}{2016}\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2014}{2015}\times\frac{2015}{2016}\)
\(=\frac{1}{2016}\)
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
A) \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)
\(x=\dfrac{22}{15}\)
b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)
C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)
Bài 1
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2
\(\dfrac{111}{333}=\dfrac{111:111}{333:111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222:2222}{4444:2222}=\dfrac{1}{2}\)
Do \(3>2\Rightarrow\dfrac{1}{3}< \dfrac{1}{2}\)
Vậy \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
Bài 1.
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98\times99\times...\times2005}{99\times100\times...2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2.
Có: \(\dfrac{111}{333}=\dfrac{111}{3\times111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222}{2\times2222}=\dfrac{1}{2}\)
Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) nên \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6}{2\cdot3\cdot4\cdot5\cdot6\cdot7}\)
\(=\frac{1}{7}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times...\times\left(1-\dfrac{1}{2015}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times...\times\dfrac{2014}{2015}\\ =\dfrac{1}{2015}\)
1/2015