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a/ Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)
còn sao nx thì mk chịu =.=
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)
a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left(\dfrac{-5}{9}+\dfrac{-13}{9}\right)\)
\(=\dfrac{3}{11}.\left(-2\right)\)
\(=\dfrac{-6}{11}\)
b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)
\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)
\(=\dfrac{77}{9}-\dfrac{9}{5}\)
\(=\dfrac{385}{45}-\dfrac{81}{45}\)
\(=\dfrac{304}{45}\)
c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{696}{261}\)
\(=-\dfrac{692}{261}\)
d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=0+0+0+4-1-1-1\)
\(=4-3\)
\(=1\)
\(a)\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)-\left(\dfrac{5}{7}-\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{8}{18}\right)-\left(\dfrac{10}{14}-\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}-\dfrac{9}{18}-\left(-\dfrac{7}{14}\right)\)
\(=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b)1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=0+0+0-\dfrac{2}{2}-\dfrac{3}{3}-\dfrac{4}{4}\)
\(=0-1-1-1\)
\(=-3\)
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)
= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)
= 0 + \(\dfrac{11}{125}\)
= \(\dfrac{11}{125}\)
b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +
\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4
= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4
= -1
c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)
= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)
= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)
= \(\dfrac{-7}{75}\)
d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)
= 1 + (-1)
= 0
= 3 - 4 - 1 - 1 - 3/4
= -15/4
\(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}-4-\dfrac{1}{3}-2-\dfrac{1}{2}-1=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(3-4\right)-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(2-2\right)-\dfrac{3}{4}=0-1-1-1+0-\dfrac{3}{4}=-3-\dfrac{3}{4}=-\dfrac{15}{4}\)