Tìm x biết: \(\left(1+\dfrac{1}{1.3}\right)\cdot\left(1+\dfrac{1}{2.4}\right)\cdot...\cdot\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{2.2017}{2018}\)
Những câu hỏi liên quan
1.Tính A= \(\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\cdot\cdot\left(\dfrac{1}{2018}-1\right)\)
2. Tìm GTLN của B = \(-\left|2018\cdot x+1\right|+\dfrac{3}{13}\)
1,
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)
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1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
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tìm x:
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
Ta có:
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{10}\right)=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3.....9}{2.3.4.....10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
\(\Leftrightarrow x=201\)
Vậy x = 201
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\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{10}\right)=\)\(\dfrac{x}{2010}\)
\(\Leftrightarrow\) \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}......\dfrac{9}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1.2.3......9}{2.3.4.......10}=\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{1}{10}=\dfrac{x}{2010}\)
\(\Leftrightarrow2010=10x\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
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1: rút gọn rồi tính
left(-dfrac{72}{40}-dfrac{144}{60}-2dfrac{1}{3}right) : left(dfrac{45}{100}-dfrac{25}{60}+-dfrac{75}{25}right)
2: tìm x: 3cdotleft(4-xright)+left(x+2right)cdotleft(1+2xright)7cdotleft(1+xright)-2xcdotleft(2-xright)
3: tìm x: dfrac{2cdotleft(1+xright)}{3}-dfrac{5cdotleft(2-xright)}{6}1dfrac{1}{3}-dfrac{3cdotleft(2x+3right)}{4}-1dfrac{1}{2}cdotleft(x+1right)
4: cho a 3+3^{2^3}+3^3+3^4+...+3^{360}
Đọc tiếp
1: rút gọn rồi tính
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right)\) : \(\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
2: tìm x: \(3\cdot\left(4-x\right)+\left(x+2\right)\cdot\left(1+2x\right)=7\cdot\left(1+x\right)-2x\cdot\left(2-x\right)\)
3: tìm x: \(\dfrac{2\cdot\left(1+x\right)}{3}-\dfrac{5\cdot\left(2-x\right)}{6}=1\dfrac{1}{3}-\dfrac{3\cdot\left(2x+3\right)}{4}-1\dfrac{1}{2}\cdot\left(x+1\right)\)
4: cho a= \(3+3^{2^3}+3^3+3^4+...+3^{360}\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
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Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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Xem thêm câu trả lời
1/Sleft(1+dfrac{1}{2}right)cdotleft(1+dfrac{1}{3}right)cdotleft(1+dfrac{1}{4}right)cdot...cdotleft(1+dfrac{1}{100}right)
2/Bleft(1-dfrac{1}{2}right)cdotleft(1-dfrac{1}{3}right)cdotleft(1-dfrac{1}{4}right)cdot...cdotleft(1-dfrac{1}{2007}right)
3/Cdfrac{2^2}{1cdot3}cdotdfrac{3^2}{2cdot4}cdotdfrac{4^2}{3cdot5}cdot...cdotdfrac{100^2}{99cdot101}
Đọc tiếp
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
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Tính GTCBT
\(A=\dfrac{4x^4+1}{4\left(x+1\right)^2}\cdot\dfrac{4\left(x+2\right)^2+1}{4\left(x+3\right)^4+1}\cdot\cdot\cdot\dfrac{4\left(x+10\right)^4+1}{4\left(x+11\right)^4+1}\)
Tại x =19,092014
Tính:
\(D=\dfrac{4x^2-1}{\left(x-y\right)\cdot\left(x+y\right)}+\dfrac{4y^2-1}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{4z^2-1}{\left(z-x\right)\cdot\left(z-y\right)}\)