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\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)
\(\Rightarrow B=1+\dfrac{1}{2}.2.3\div2+\dfrac{1}{3}.3.4\div2+...+\dfrac{1}{20}.20.21\div2\)
\(\Rightarrow B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)
\(\Rightarrow B=\dfrac{2+3+4+...+21}{2}\)
\(\Rightarrow B=\dfrac{230}{2}\)
\(\Rightarrow B=115\)
Vậy \(B=115\)
\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{20}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\right)\)
\(A=1-\dfrac{1}{2^{20}}=\dfrac{2^{20}-1}{2^{20}}\)
Chọn A
Ta thấy:
\(1+2=\dfrac{2\cdot\left(2+1\right)}{2}\\ 1+2+3=\dfrac{3\cdot\left(3+1\right)}{2}\\ 1+2+3+4=\dfrac{4\cdot\left(4+1\right)}{2}\\ ...\\ 1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\\ =1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\\ =1+\dfrac{1\cdot2\cdot3}{2\cdot2}+\dfrac{1\cdot3\cdot4}{3\cdot2}+...+\dfrac{1\cdot20\cdot21}{20\cdot2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\\ =\dfrac{2+3+4+...+21}{2}\\ =\dfrac{1+2+3+..+21-1}{2}\\ =\dfrac{\left(\dfrac{21\cdot22}{2}\right)-1}{2}\\ =\dfrac{231-1}{2}\\ =\dfrac{230}{2}\\ =115\)
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
\((\dfrac{1}{2})^{15}\times(\dfrac{1}{2})^{20}=(\dfrac{1}{2})^{15+20}=(\dfrac{1}{2})^{35}\) \([(\dfrac{1}{3})^2]^{25}\div(\dfrac{1}{3})^{30}=(\dfrac{1}{3})^{50}\div(\dfrac{1}{3})^{30}=(\dfrac{1}{3})^{50-30}=(\dfrac{1}{3})^{20}\) \((\dfrac{1}{16})^3\div(\dfrac{1}{8})^2=[(\dfrac{1}{2})^4]^3\div[(\dfrac{1}{2})^3]^2=(\dfrac{1}{2})^{12}\div(\dfrac{1}{2})^6=(\dfrac{1}{2})^{12-6}=(\dfrac{1}{2})^6\) (x^3)^2 : ( x^2)^3= x^6 :x^6=1
\(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{20}=\left(\dfrac{1}{2}\right)^{15+20}=\left(\dfrac{1}{2}\right)^{35}\)
\(\left(\dfrac{1}{9}\right)^{25}:\left(\dfrac{1}{3}\right)^{30}=\left[\left(\dfrac{1}{3}\right)^2\right]^{25}:\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{50}:\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{50-30}=\left(\dfrac{1}{30}\right)^{20}\)\(\left(\dfrac{1}{16}\right)^3:\left(\dfrac{1}{8}\right)^2=\left[\left(\dfrac{1}{2}\right)^4\right]^3:\left[\left(\dfrac{1}{2}\right)^3\right]^2=\left(\dfrac{1}{2}\right)^{12}:\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^{12-6}=\left(\dfrac{1}{2}\right)^6\)
\(\left(x^3\right)^2:\left(x^2\right)^3=x^6:x^6=x^0=1\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
\(=0\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)