\(\left(1_3^1\right)\left(1_8^1\right)\left(1_{15}^1\rig...">
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\(A=1\dfrac{1}{3}\cdot1\dfrac{1}{8}\cdot...\cdot1\dfrac{1}{99}\)

\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot...\cdot\dfrac{100}{99}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{10\cdot10}{9\cdot11}\)

\(=\dfrac{2\cdot3\cdot...\cdot10}{1\cdot2\cdot...\cdot9}\cdot\dfrac{2\cdot3\cdot...\cdot10}{3\cdot4\cdot...\cdot11}\)

\(=\dfrac{10}{1}\cdot\dfrac{2}{11}=\dfrac{20}{11}\)

17 tháng 1

Đặt A = 4/3 * 9/8 * 16/15 *...*100/99

-> A = (2*2*3*3*4*4...*10*10)/(1*3*2*4*3*5*...*9*11)

-> A = (2*10)/(11*1) = 20/11
Vạy phép tính = 20/11



a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)

\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)

\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)

\(=-\frac{19}{8}\)

b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)

\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)

\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)

\(=\frac{5}{12}\)

c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)

\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)

\(=-\frac{43}{30}\)

d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)

\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)

\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)

\(=\frac{3}{20}\)

e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)

\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)

f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)

\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)

g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)

\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)

\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)

h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)

\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)

\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)

\(=\frac{78149}{6210}\)

i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)

\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)

\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)

\(=\frac{55}{4}\)

11 tháng 3 2017

toàn hỏi lung tung. lớp 6 mà còn ko biết làm mấy bài toán vớ vẩn kia

3 tháng 5 2019

\(B=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(B=\frac{1}{3}.\frac{102}{103}\)

\(B=\frac{34}{103}\)

3 tháng 5 2019

Bài 3: đổi ra phân số rồi tính, đổi:\(1,5=\frac{15}{10};2,5=\frac{25}{10};1\frac{3}{4}=\frac{7}{12}\)(cái này ko giải dùm, đổi ra như thek rồi tính nha)

\(B=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{3}.\frac{102}{103}\)

\(=\frac{1}{1}.\frac{34}{103}=\frac{34}{103}\)

8 tháng 9 2017

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)

7 tháng 10 2017

\(=45-\left(3^2+2^4\right)=45-\left(9+16\right)=45-25=20\)

7 tháng 10 2017

b)\(=50+\left(30-2\left(14-3\right)\right)=50+\left(30-22\right)=50+8=58\)

16 tháng 4 2017

Sách Giáo Khoa

Lời giải

a) 15.(-2).(-5).(-6)
 = [15.(-2)].[(-5)].(-6)
 = -30.30
 = -900

hoc
   15.(-2).(-5).(-6)
 = [15.(-6)].[(-2).(-5)]
 = -90.10
 = -900
 
b) 4.7.(-11).(-2)
 = (4.7).[(-11).(-2)]
 = 28.22
 = 616

a) 15.(-2).(-5).(-6)

= [15.(-2)].[(-5)].(-6)

= -30.30

= -900

hoặc 15.(-2).(-5).(-6)

= [15.(-6)].[(-2).(-5)]

= -90.10

= -900

b) 4.7.(-11).(-2)

= (4.7).[(-11).(-2)]

= 28.22

= 616

29 tháng 8 2018

a) 2011 + 5[300 - (17 - 7)2 ]

= 2011 + 5[300 - 100]

= 2011 + 5.200

= 2011 + 1000

= 3011

b) 695 - [200 + (11 - 1)2 ]

= 695 - [200 + 100]

= 695 - 300

= 395

c) 129 - 5[29 - (6-1)]

= 129 - 5[29 - 25]

= 129 - 5 . 4

= 129 - 20 

= 109

d) 2345 - 1000 : [19 - 2(21-18)2 ]

= 2345 - 1000 : [19 - 2 . 9]

= 2345 - 1000 : 1

= 2345 - 1000

= 1345

29 tháng 8 2018

a, 2011+5 [300-(17-7)2 ]  =2011+5[300-102

=2011+5[300-100]=2011+5.200

=2011+1000=3011

b, 695-[200+(11-1)2]=695-[200+102]

=695-[200+100]=695-300=395