Tìm x:
a) \(x-\left(\frac{50x}{160}+\frac{25x}{200}\right)=11\frac{1}{4}\)
b)\(\left(x-5\right)\cdot\frac{30}{100}=\frac{200x}{100}+5\)
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1 tháng 7 2018
\(\frac{2}{2.3}\) + \(\frac{2}{3.4}\) + \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + .......+ \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{2019}\) : 2
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2}\) - \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2019}\)
<=> x + 1 = 2019 => x = 2018
vậy x = 2018
1 tháng 7 2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
31 tháng 3 2020
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
1 tháng 8 2018
\(c,\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow(x+1)(x+1)=2.8\)
\(\Rightarrow(x+1)^2=16\)
\(\Rightarrow(x+1)^2=4^2\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
1 tháng 8 2018
\(a,x-(\frac{50x}{100}+\frac{25x}{200})=11\frac{1}{4}\)
\(\Rightarrow x-\frac{50x+25x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x}{100}-\frac{75x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x-75x}{100}=\frac{1125}{100}\)
\(\Rightarrow25x=1125\)
\(\Rightarrow x=45\)
29 tháng 6 2018
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2018}\)= \(\frac{2017}{2018}\)< 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 ( dpcm )
29 tháng 6 2018
Ta có:
\(\frac{1}{2^2}\)< \(\frac{1}{1.2}\).
\(\frac{1}{3^2}\)< \(\frac{1}{2.3}\).
\(\frac{1}{4^2}\)< \(\frac{1}{3.4}\).
...
\(\frac{1}{2017^2}\)< \(\frac{1}{2016.2017}\).
\(\frac{1}{2018^2}\)< \(\frac{1}{2017.2018}\).
Từ trên ta có:
\(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+...+ \(\frac{1}{2016.2017}\)+ \(\frac{1}{2017.2018}\)= 1- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{2016}\)- \(\frac{1}{2017}\)+ \(\frac{1}{2017}\)- \(\frac{1}{2018}\)= 1- \(\frac{1}{2018}\)< 1.
=> \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< 1.
=> ĐPCM.
13 tháng 3 2019
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
tui o bít nhưng ai kb vs tui o