2019!-2018!-\(2018^2\)! bằng ?
Những câu hỏi liên quan
11 tháng 2 2021 lúc 21:53
So sánh:
\(C=\dfrac{2019-2018}{2018+2019}\) và \(D=\dfrac{2019^2-2018^2}{2019^2+2018^2}\)
Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)
\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)
\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)
Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)
\(2019^2+2018^2=2019^2+2018^2+0\)
Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)
\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)
\(\Leftrightarrow C< D\)
Đúng 0
Bình luận (0)
Tính bằng cách thuận tiện nhất 2018/2019 x 9/13 x 5/13 x 2018/2019 - 2018/2019 x 1/13
so sánh A=2018^2019 -1/2018^2019+1 và B = 2018^2019/2018^2019+2
Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A > B
SO SÁNH
A=2018^2019-1/2018^2019+1 VÀ B =2018^2019/2018^2019+2
\(A=\frac{2018^{2019}-1}{2018^{2019}+1}=\frac{2018^{2019}+1-2}{2018^{2019}+1}=\frac{2018^{2019}+1}{2018^{2019}+1}-\frac{2}{2018^{2019}+1}=1-\frac{2}{2018^{2019}+1}\)
\(B=\frac{2018^{2019}}{2018^{2019}+2}=\frac{2018^{2019}+2-2}{2018^{2019}+2}=\frac{2018^{2019}+2}{2018^{2019}+2}-\frac{2}{2018^{2019}+2}=1-\frac{2}{2018^{2019}+2}\)
Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)
\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)
\(\Rightarrow A< B\)
Vậy .....
Đúng 0
Bình luận (0)
cho x^2018+y^2018+z^20018+t^2018/a^2+b^2+c^2+d^2
=x^2018/a^2+y^2018/b^2+z^2018/c^2+t^2018/d^2tính T=x^2019+y^2019+z^2019+t^2019
giúp mik nha mn ơi.
mik cần gấp bâgiowf
Cho
A= 20172018 +10/ 20182019 +10
B= 20182019 + 10/ 20192020 +10
So sánh A và B (Bằng 2 cách)
Cho M=(2018^2018+2019^2018)^2019 và N=(2018^2019+2019^2019)^2018. So sánh M và N
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]