\(\left\{\left[\left(6.2:0.31-\frac{5}{6}\cdot0.9\right)\cdot0.2+0.15\right]:0.2\right\}:\left[\left(2+1\frac{4}{11}\cdot0.22:0.1\right)\cdot\frac{1}{33}\right]\)
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
\(\left[6\cdot\left(-\frac{1}{3}\right)^2-3\cdot\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(=\left[6\cdot\left(-\frac{1}{9}\right)+1+1\right]:\left(-\frac{4}{3}\right)\)
\(=\left(-\frac{2}{3}+2\right):\left(-\frac{4}{3}\right)\)
\(=\frac{4}{3}:\left(-\frac{4}{3}\right)=-1\)
Tính
\(3\frac{1}{2}\cdot\frac{4}{49}\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\left(\frac{-42}{5}\right)\)
\(\left[0,\left(5\right).0,\left(2\right)\right]:\left(3\frac{1}{3}:\frac{33}{25}\right)-\left(\frac{2}{5}\cdot1\frac{1}{3}\right):\frac{4}{3}\)
Bài 1: Tính a) \(\left(\frac{11}{12}:\frac{44}{16}\right)\cdot\left(\frac{-1}{3}+\frac{1}{2}\right)\) b) \(\frac{\left(-5^2\right)\cdot\left(-5\right)^3\cdot16}{5^4\cdot\left(-2\right)^4}\) c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\)d) \(\left(\frac{-1}{2}+\frac{1}{3}\right)\cdot\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{4}{5}\)
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
TÌM x
\(\left(\left(\frac{3}{4}\cdot x+5\right)-\left(\frac{2}{3}\cdot x-4\right)-\left(\frac{1}{6}\cdot x+1\right)\right)=\left(\frac{1}{3}\cdot x+4\right)-\left(\frac{1}{3}-3\right)\)
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
TÍNH
\(C=\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{2}{2015}\right)\cdot\left(1+\frac{2}{2017}\right)\)
\(D=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\left(1-\frac{1}{10}\right)\cdot\left(1-\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
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:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
p/s: chúc bạn hok tốt
Tính:
a)\(\left\{\left[\left(6,2:0,31-\frac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\frac{4}{11}:0,1\right).\frac{1}{33}\right]\)
b)\(0,4\left(3\right)+0,6\left(2\right)-2\frac{1}{2}.\left[\left(\frac{1}{2}+\frac{1}{3}:0,5\left(8\right)\right)\right]:\frac{50}{53}\)
c)\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)