thực hiện các phép tính
a) \(-1\dfrac{1}{2}.21\dfrac{1}{3}+1\dfrac{1}{2}.1\dfrac{1}{3}\)
b) \(\dfrac{2018}{2019}.\left(13-13\dfrac{2018}{2019}\right)-\dfrac{1}{2019}:\dfrac{2019}{2018}\)
A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)
so sánh A và B biết:
A=\(\dfrac{2^{2018}}{2^{2018}+3^{2019}}\)+\(\dfrac{3^{2019}}{3^{2019}+5^{2020}}\)+\(\dfrac{5^{2020}}{5^{2020}+2^{2018}}\)
B=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{2019.2020}\).
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
Tính A/B
A=
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)
B=
\(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
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) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)
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: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
Giải:
Ta có:
\(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
\(B=1+\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)\)
\(B=\dfrac{2021}{2021}+\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}\)
\(B=2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\left[2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\right]}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=2021\)
Vậy \(\dfrac{A}{B}=2021\)
Bài 1: Tính giá trị của biểu thức sau
A=1-\(\dfrac{50-\dfrac{4}{2018}+\dfrac{2}{2019}-\dfrac{2}{2020}}{100-\dfrac{8}{2018} +\dfrac{4}{2019}-\dfrac{4}{2020}}\)
B=\(\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
C=\(x^{2020}\)-\(y^{2020}\)+\(xy^{2019}\)-\(x^{2019}\).y+2019 biết x-y=0
Mong mn giúp đỡ
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
rút gon biểu thức:
1, \(\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\dfrac{1}{a^2}-1}-\dfrac{1}{a}\right)\)
2, \(\dfrac{1+2019\sqrt{2018}-2018\sqrt{2019}}{\sqrt{2018}+\sqrt{2019}+\sqrt{2018.2019}}\)
1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
Kết quả phép tính:\(\dfrac{2018}{2019}\)-\(\dfrac{2017}{2018}\)
a.1/2018 b.1/2019x2018 c.1 d.1/2019
Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
Đề bài: So sánh
1, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}với\) 3
2, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}với\dfrac{2017+2018}{2018+2019}\)
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Bài 6: So sánh
a,\(\dfrac{1}{2}\)+\(\dfrac{1}{_{ }2^2}\)+\(\dfrac{1}{2_{ }^3}\)+...+\(\dfrac{1}{2^{2014}}\)và 1 b,\(\dfrac{10^{2018}+5}{10^{2018}-8}\)và \(\dfrac{10^{2019}+5}{10^{2019}-8}\)
c,\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{23.24.25}\)và\(\dfrac{1}{4}\)