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26 tháng 7 2017

A = (1 - 1/2) x (1 - 1/3) x (1 - 1/4) + ... + (1 - 1/2017) x (1 - 1/2018)
<=> A = 1/2 x 2/3 x 3/4 x ... x 2016/2017 x 2017/2018
<=> A = \(\dfrac{1\times2\times3\times...\times2016\times2017}{2\times3\times4\times...\times2017\times2018}\)
<=> A = \(\dfrac{1}{2018}\)
@Nguyễn Linh Ly

26 tháng 7 2017

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\left(1-\dfrac{1}{2018}\right)\)\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2016}{2017}.\dfrac{2017}{2018}\)

\(A=\dfrac{1.2.3....2016.2017}{2.3.4....2017.2018}\)

\(A=\dfrac{1}{2018}\)

31 tháng 7 2018

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2017}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\)

\(A=\frac{1}{2017}\)

31 tháng 7 2018

\(\frac{1-1}{2}.\frac{1-1}{3}.\frac{1-1}{4}......\frac{1-1}{2017}.\frac{1-1}{2018}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}........\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1}{2018}\)

12 tháng 5 2018

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow x+2=41\)

\(\Leftrightarrow x=41-2\)

\(\Leftrightarrow x=39\)

5 tháng 4 2020

???????????????????????????????????????????????????????

7 tháng 8 2018

help me

7 tháng 8 2018

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

24 tháng 7 2017

=\(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{2016}{2017}.\dfrac{2017}{2018}\)

\(\dfrac{1.2.3.4....2017}{2.3.4....2017.2018}\)

=\(\dfrac{1}{2018}\)

25 tháng 7 2017

Thanks,Linhokeoeo

8 tháng 7 2018

A = 1 + 2 + 3 + ... + 2018

= ( 1 + 2018 ) + ( 2 + 2017) + ... + ( 1009 + 1010 )

= 2019 + 2019 + ... + 2019 ( có 1009 số 2019 )

= 2019 x 1009 = 2037171

B = 1 + 3 + 5 + ... + 2017

= ( 1 + 2017 ) + ( 3 + 2015 ) + ... + ( 1007 + 1010) + 1009 

= 2018 + 2018 + ... + 2018 + 1009 (có 504 số 2018)

= 2018 x 504 + 1009 = 1018081

Còn lại làm giống ý trên . 

6 tháng 7 2018

\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)

\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)

4 tháng 6 2021

khó quá bẹn gì đấy ơi

28 tháng 1 2020

Ta có : A =\(\frac{2017}{2018}\)\(\frac{7}{8}\)\(\frac{2017}{2018}\)\(\frac{3}{8}\)\(\frac{2017}{2018}\)\(\frac{1}{4}\)

                = \(\frac{2017}{2018}\) x ( \(\frac{7}{8}+\frac{3}{8}-\frac{1}{4}\))

                  = \(\frac{2017}{2018}\)x 1

                    =\(\frac{2017}{2018}\)

Vậy A= : \(\frac{2017}{2018}\)

28 tháng 1 2020

                                                                   Bài giải

\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)

\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)

2 tháng 9 2019

=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2

=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)

Với x+2020=0=>x=-2020

Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí 

Vậy x=-2020