F=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......... +\dfrac{1}{2^{2018}}\).Tính F
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18 tháng 2 2022
sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Cộng vế với vế
\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4
Vậy ta có đpcm
27 tháng 8 2021
c: Ở hai hàm số trên, nếu lấy biến x cùng một giá trị thì f(x) sẽ nhỏ hơn g(x) 3 đơn vị
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2021
a. ĐKXĐ: \(x\ge-1\)
\(y=\sqrt{x^3+1+2\sqrt{x^3+1}+1}+\sqrt{x^3+1-2\sqrt{x^3+1}+1}\)
\(=\sqrt{\left(\sqrt{x^3+1}+1\right)^2}+\sqrt{\left(\sqrt{x^3+1}-1\right)^2}\)
\(=\left|\sqrt{x^3+1}+1\right|+\left|1-\sqrt{x^3+1}\right|\ge\left|\sqrt{x^3+1}+1+1-\sqrt{x^3+1}\right|=2\)
b.
\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
c.
\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)
AH
Akai Haruma
Giáo viên
12 tháng 8 2021
Lời giải:
Gọi biểu thức trên là $A$
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2018.2019}\)
\(=2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019})\)
\(=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{2018}-\frac{1}{2019})\)
\(=2(\frac{1}{2}-\frac{1}{2019})=\frac{2017}{2019}\)
SI
29 tháng 6 2021
Ta có :
B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)
B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\) (1)
Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)
29 tháng 6 2021
Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)
Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 12 2020
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
28 tháng 1 2022
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
\(F=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2018}}\)
\(\dfrac{1}{2}F=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2019}}\)
\(F-\dfrac{1}{2}F=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2018}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{2019}}\)
\(\dfrac{1}{2}F=\dfrac{1}{2}-\dfrac{1}{2^{2019}}\)
\(F=\left(\dfrac{1}{2}-\dfrac{1}{2^{2019}}\right):\dfrac{1}{2}\)
\(F=\left(\dfrac{1}{2}-\dfrac{1}{2^{2019}}\right).2\)
\(F=1-\dfrac{1}{2^{2018}}\)
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