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b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)
=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)
=\(-\frac{11}{4}\cdot\frac{8}{33}\)
=\(-\frac{2}{3}\)
b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)
=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)
=\(\frac{-1}{11}\cdot55=-5\)
c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)
=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)
=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)
=\(-\frac{2}{3}\cdot\frac{7}{5}\)
=\(\frac{-14}{15}\)
d) chưa nghĩ ra nhé
e) bạn chép sai đề bài rồi
mk mới kiểm tra 45 phút nên biết
đề bài nè
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)
=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)
=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)
=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)
=\(\frac{1.31}{2.30}\)
=\(\frac{31}{60}\)
a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong b)bn tu lam nhe c)dat thua so chung d)tinh trong ngoac ra rui nhan vs e) mk bo tay
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)
\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Rightarrow1488x-476,5x=20230\)
\(\Rightarrow1011,5x=20230\)
\(\Rightarrow x=20\)
Bài làm :
Ta có :
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)
\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Leftrightarrow1488x-476,5x=20230\)
\(\Leftrightarrow1011,5x=20230\)
\(\Leftrightarrow x=20\)
Vậy x=20