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NV
4 tháng 10 2021

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

\(3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3B-B=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow2B=1-\dfrac{1}{3^{100}}\)

\(0< \dfrac{1}{3^{100}}< 1\Rightarrow0< 1-\dfrac{1}{3^{100}}< 1\)

\(\Rightarrow0< 2B< 1\Rightarrow0< B< \dfrac{1}{2}\Rightarrow\) B không phải số nguyên

22 tháng 2 2023

a)

`(2x-1)(x+2/3)=0`

\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b)

\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)

\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)

\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)

\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)

\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)

22 tháng 2 2023

sai rồi , x không thể có 2 giá trị

30 tháng 12 2017

\(A=1-\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{4}\right)^3+...+\left(\dfrac{3}{4}\right)^{2016}-\left(\dfrac{3}{4}\right)^{2017}\\ \Rightarrow\dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+\left(\dfrac{3}{4}\right)^3-\left(\dfrac{3}{4}\right)^3+...+\left(\dfrac{3}{4}\right)^{2017}-\left(\dfrac{3}{4}\right)^{2018}\\ \Rightarrow\dfrac{7}{4}A=1-\left(\dfrac{3}{4}\right)^{2018}\notin Z\\ \Rightarrow A\notin Z\)

10 tháng 12 2022

Câu 2

(a+3)(b-4)-(a-3)(b+4)=0

=>ab-4a+3b-12-ab-4a+3b+12=0

=>-8a=-6b

=>a/b=3/4

=>a/3=b/4

\(\dfrac{2}{3}A=\dfrac{2}{3}-\left(\dfrac{2}{3}\right)^2+\left(\dfrac{2}{3}\right)^3-...+\left(\dfrac{2}{3}\right)^{2019}-\left(\dfrac{2}{3}\right)^{2020}\)

=>\(\dfrac{5}{3}A=1-\left(\dfrac{2}{3}\right)^{2020}=1-\dfrac{2^{2020}}{3^{2020}}=\dfrac{3^{2020}-2^{2020}}{3^{2020}}\)

=>\(A=\dfrac{3^{2020}-2^{2020}}{3^{2020}}:\dfrac{5}{3}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên

11 tháng 10 2023

a) = 0,16 - 0,064 . (-3)

= 0,16 + 0,192

= 0,352

b) = (7/4)³ - (7/4)² + 1

= 343/64 - 49/16 + 1

= 147/64 + 1

= 211/64

c) = 8/27 - 4.(-7/4)² - 8/27

= -4.49/16

= -49/4

AH
Akai Haruma
Giáo viên
10 tháng 10 2023

Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$

$=\frac{147}{64}+1=\frac{211}{64}$

c.

$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$

$=-4(\frac{-7}{4})^2=\frac{-49}{4}$