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=(2015/ 2019 + 3/2019 + 1/2019 ) : 1/2
= 2019/2019 x 2
= 1 x2
=2
2015/2019:1/2+3/2019:1/2+1/2019:1/2
=(2015/2019+3/2019+1/2019):1/2
=1:1/2
=2
k cho mink nha
ta có 1/2*2/3*...*2019/2020
=1*2*3*...*2019/2*3*4*..*2020
=1/2020 (rút gọn các số giống nhau)
Ok em, để olm.vn giúp em nhá:
A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020
A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1009}{2020}\)
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
1) 2*17*9+18*540+29*18
= 18*17+18*540+29*18
= 18*(17+540+29)
= 18*586
= 10548
2) 5*{26-[3*(5+2*5)+15]/15}
= 5*{26-[3*(5+10)+15]/15}
= 5*{26-[3*15+15]/15}
= 5*{26-[45+15]/15}
= 5*{26-60/15}
= 5*{26-4}
=5*22
=110
3) (2018*2019+2019*2020)*(45*120-15*360)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*(15*3*120-15*120*3)*(1+5+9+13+17+...+2015+2019)
= (2018*2019+2019*2020)*0*(1+5+9+13+17+...+2015+2019)
= 0
số lượng số hạng của dãy số là
( 2021 - 2 ) : 1 + 1 = 2020
tổng của dãy số là
( 2021 + 2) x 2020 : 2 = 2043230
vậy A = \(\frac{1}{2043230}\)
Ta có:\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
Vậy biểu thức \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020
= 1 x 2 x 3 x ... x 2018 x 2019 / 2 x 3 x 4 x ... x 2019 x 2020
Khử loại đi ta còn lại phân số 1/2020
Hok tốt ^^
\(\left(y+2\dfrac{1}{3}\right)+6\%=\left(2\dfrac{1}{3}+6\%\right)+2019\)
\(\left(y+\dfrac{7}{3}\right)+\dfrac{3}{50}=\left(\dfrac{7}{3}+\dfrac{3}{50}\right)+2019\)
\(\left(y+\dfrac{7}{3}\right)+\dfrac{3}{50}=\dfrac{359}{150}+2019\)
\(\left(y+\dfrac{7}{3}\right)+\dfrac{3}{50}=\dfrac{303209}{150}\)
\(y+\dfrac{7}{3}=\dfrac{303209}{150}-\dfrac{3}{50}\)
\(y+\dfrac{7}{3}=\dfrac{6064}{3}\)
\(y=\dfrac{6064}{3}-\dfrac{7}{3}\)
\(y=2019\)
\(\text{Cách 2}\) \(\left(y+2\dfrac{1}{3}\right)+6\%=\left(2\dfrac{1}{3}+6\%\right)+2019\)
\(\left(y+\dfrac{7}{3}\right)+\dfrac{3}{50}=\left(\dfrac{7}{3}+\dfrac{3}{50}\right)+2019\)
\(y+\left(\dfrac{7}{3}+\dfrac{3}{50}\right)=\left(\dfrac{7}{3}+\dfrac{3}{50}\right)+2019\)
\(y-2019=\left(\dfrac{7}{3}+\dfrac{3}{50}\right)-\left(\dfrac{7}{3}+\dfrac{3}{50}\right)\)
\(y-2019=0\)
\(y=0+2019\)
\(y=2019\)
\(\frac{1}{1}+2+\frac{1}{1}+2+3+\frac{1}{1} +2+3+4+...+\frac{1}{1}+2+3......2019\)
Ta có : \(\frac{2}{2}+\left(\frac{1}{2}\right)+\frac{2}{2}+\left(1+2+3\right)+....+\frac{2}{2}+\left(1+2+....+50\right)\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{2550}\)
\(=\frac{2}{2}.3+\frac{2}{3}.4+....+\frac{2}{50}.51\)
\(=2.\left(\frac{1}{2}.3+\frac{1}{3}.4+.....+\frac{1}{50}.51\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2.\left(1-\frac{1}{51}\right)\)
\(=2.\frac{50}{51}\)
\(=\frac{100}{51}\)
Hmmm , kh bt có đúng kh nhỉ ???
Nếu kh đúng chỗ nào mong m.n chỉ ạ
:>>
số số hạng của tổng trên lak:
(2019 - 1) : 1 + 1 = 2019 (số hạng )
tổng trên có giá trị lak :
(2019 + 1 ) . 2019 : 2 =2039190
và bn tự đáp số thui
đúng thì k mk nhoa\\\\
thanks nhìu ak##
số số hạng:(2019-1)+1=2019
tổng :(2019+1)x2019/2=...(bn tự tính)
hok tốt