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Câu b
Ta có :x + 3 /1.3 +3/3.5 + 3/5.7+...+3/13.15=2 1/5
X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1=11/5
X+2/3.(1-1/15)=11/5
X+ 2/3.14/15=11/5
X + 28/45=11/5
X = 11/5 -28/45
X=71/45
Câu a gợi ý
1/2-1/3/1/6=0
1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) =0
3/4:x=9/10
X = 3/4:9/10
X = 5/6
a) (x - 1) + (x + 2) + (x - 3) + (x + 4) + (x - 5) + (x + 6) = 21
x - 1 + x + 2 + x - 3 + x + 4 + x - 5 + x + 6 = 21
6x + 3 = 21
6x = 21 - 3
6x = 18
x = 18 : 6
x = 3
b) (x - 1) + (x + 3) + (x - 5) + (x + 7) + (x - 9) + (x + 11) = 186
x - 1 + x + 3 + x - 5 + x + 7 + x - 9 + x + 11 = 186
6x + 6 = 186
6x = 186 - 6
6x = 180
x = 180 : 6
x = 30
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right):2}=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right)}\times\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4016}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)
a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)
\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
\(\left(1+\dfrac{1}{2010}\right)\times\left(1+\dfrac{1}{2011}\right)\times...\times\left(1+\dfrac{1}{2020}\right)\)
=\(\dfrac{2011}{2010}\times\dfrac{2012}{2011}\times...\times\dfrac{2021}{2020}\)
=\(\dfrac{2021}{2010}\)
\(\dfrac{2020x2018+9}{2019x2020-2011}=\dfrac{2020x\left(2019-1\right)+9}{2020x2019-2011}=\dfrac{2020x2019-2020+9}{2020x2019-2011}=\dfrac{2020x2019-2011}{2020x2019-2011}=1\)\(\left(1+\dfrac{1}{2}\right)x\left(1+\dfrac{1}{3}\right)x\left(1+\dfrac{1}{4}\right)x...x\left(1+\dfrac{1}{2020}\right)\\ =\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x...x\dfrac{2021}{2020}=\dfrac{2021}{2}\)