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E=\(\frac{1}{5}\).(\(\frac{1}{11}-\frac{1}{16}\)+\(\frac{1}{16}-\frac{1}{21}+\frac{1}{21}+\frac{1}{26}+....+\frac{1}{61}-\frac{1}{66}\))
E=\(\frac{1}{5}.\left(\frac{1}{11}-\frac{1}{66}\right)\)=\(\frac{1}{5}.\frac{5}{66}=\frac{1}{66}\)
\(E=\frac{1}{11x16}+\frac{1}{16x21}+\frac{1}{21x26}+...+\frac{1}{61x66}\)
\(E=\frac{1}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}+\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\right)\)
\(E=\frac{1}{5}\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(E=\frac{1}{5}.\frac{5}{66}\)
\(E=\frac{1}{66}\)
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{9}\right)+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{2}.\frac{8}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{4}{9}+\frac{1}{x}=1\)
\(\Rightarrow\frac{1}{x}=1-\frac{4}{9}\)
\(\Rightarrow\frac{1}{x}=\frac{5}{9}\)
\(\Rightarrow x=\frac{1.9}{5}\)
\(\Rightarrow x=\frac{9}{5}\)
Vậy x = \(\frac{9}{5}\)
b) \(\frac{2}{3}-\frac{1}{3}.\left(x-2\right)=\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{2}{3}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}.\left(x-2\right)=\frac{5}{12}\)
\(\Rightarrow x-2=\frac{5}{12}:\frac{1}{3}\)
\(\Rightarrow x-2=\frac{5}{4}\)
\(\Rightarrow x=\frac{5}{4}+2\)
\(\Rightarrow x=\frac{13}{4}\)
Vậy x = \(\frac{13}{4}\)
_Chúc bạn học tốt_
a=1/1x2+1/2x3+....+1/99x100
a=1-1/2+1/2-1/3+....+1/99-1/100
a=1-1/100
a=99/100
b=4/1x3+4/3x5+.....+4/51x53
b=2x(2/1x3+2/3x5+....+2/51x53)
b=2x(1-1/3+1/3-1/5+...+1/51-1/53)
b=2x(1-1/53)
b=2x52/53
b=104/53
đúng tick cho mình nha
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
Ta có : \(C=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+......+\frac{2}{41.42}\)
\(C=2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{41.42}\right)\)
\(C=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{41}-\frac{1}{42}\right)\)
\(C=2\left(\frac{1}{3}-\frac{1}{42}\right)\)
\(C=2.\frac{13}{42}=\frac{13}{21}\)
\(\left(x\cdot2,4-4,2\right)\div x=1\)
\(\Rightarrow x\cdot2,4-4,2=x\)
\(\Rightarrow x\cdot2,4=x+4,2\)
\(\Rightarrow\frac{12x}{5}=\frac{5x+21}{5}\)
\(\Rightarrow12x=5x+21\)
\(\Rightarrow12x-5x=21\)
\(\Rightarrow7x=21\Rightarrow x=\frac{21}{7}=3\)
Vậy x = 3
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{x+1}{2x+2}\)
Bài 1:
A = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1010}{2021}\)
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