\(\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{3}\right)^2+2.\left(-\dfrac{1}{2}\right)^2-2021^0\)
Thực hiện phép tính:
a) 2021 - \(\left(\dfrac{1}{3}\right)^2\) . 32
b \(\dfrac{5}{10}\) + 9 . \(\dfrac{-3}{2}\)
c) -10 . \(\left(-\dfrac{2021}{2022}\right)^0\) + \(\left(\dfrac{2}{5}\right)^2\) : 2
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
Tổng \(S=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2021}\)có giá trị là
Theo kinh nghiệm của tui thì.......mấy cái bài này hay dễ ra kết quả = 1 với = 0 nhiều lắm:)
\(S=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2021}\)
\(\Rightarrow\left(-\dfrac{1}{5}\right)S=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2021}+\left(-\dfrac{1}{5}\right)^{2022}\)
\(\Rightarrow S-\left(-\dfrac{1}{5}\right)S=\left(-\dfrac{1}{5}\right)^0-\left(-\dfrac{1}{5}\right)^{2022}\)
\(\Rightarrow\dfrac{6}{5}S=1-\dfrac{1}{5^{2022}}\)
\(\Rightarrow S=\dfrac{5}{6}\left(1-\dfrac{1}{5^{2022}}\right)\)
\(\dfrac{-5}{2}-\left(\dfrac{-7}{2}-\dfrac{7}{4}\right):\left(\dfrac{-7}{3}\right)-\left(\dfrac{-7}{3}\right)-\left(2021\right)^0\)
\(\dfrac{-5}{2}-\left(\dfrac{-7}{2}-\dfrac{7}{4}\right):\left(\dfrac{-7}{3}\right)-\left(\dfrac{-7}{3}\right)-2021^0\)
\(=\dfrac{-5}{2}-\dfrac{9}{4}+\dfrac{7}{3}-1\)
\(=-\dfrac{41}{12}\)
Câu 1: Thực hiện phép tính
a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)
b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)
Câu 2: Tìm x
\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
Cho 2022 số tự nhiên a(1), a(2), a(3), ..., a(2021), a(2022) khác 0 thỏa mãn:
\(\dfrac{1}{a\left(1\right)}\) + \(\dfrac{1}{a\left(2\right)}\) + ... + \(\dfrac{1}{a\left(2021\right)}\) + \(\dfrac{1}{a\left(2022\right)}\) = 1. Chứng minh rằng: tồn tại ít nhất một số trong 2022 số đã cho là số chẵn.
Tính Q= \(\dfrac{1}{1+\left(\dfrac{1}{2021}\right)^2}+\dfrac{1}{1+\left(\dfrac{2}{2020}\right)^2}+...+\dfrac{1}{1+\left(\dfrac{2021}{1}\right)^2}\)
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
cho
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\)
\(B=\left(\dfrac{3}{2}\right)^{2013}:2\)
tính B-A
Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)
Lấy (2) - (1) ta được:
\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)
Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)
Tính A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).....\left(\dfrac{1}{2021^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2021}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{2021}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-2020}{2021}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2022}{2021}\)
\(=\dfrac{1}{2021}\cdot\dfrac{2022}{2}=\dfrac{1011}{2021}\)