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Ta có:

\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2^2A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

\(\Rightarrow4A-A=1-\frac{1}{2^{100}}< 1\Rightarrow3A< 1\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)

23 tháng 12 2018

\(2^2A=1+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(4A-A=1-\frac{1}{2^{100}}\)

\(A=\frac{1-\frac{1}{2^{100}}}{3}\)

17 tháng 9 2017

hbewjfewi

11 tháng 1 2020

Câu 3 = (5 mũ 51 - 1) : 4

AH
Akai Haruma
Giáo viên
15 tháng 9 2021

Lần sau bạn lưu ý gõ đề bằng bộ gõ công thức toán $(\sum)$ để được hỗ trợ tốt hơn.

Lời giải:
Ta có:

$\frac{1}{3^2}< \frac{1}{2.3}$

$\frac{1}{4^2}< \frac{1}{3.4}$

...........

$\frac{1}{1990^2}< \frac{1}{1989.1990}$

Cộng tất cả theo vế:

$\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{1989.1990}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{1989}-\frac{1}{1990}$

$=\frac{1}{2}-\frac{1}{1990}< \frac{1}{2}$

$\Rightarrow \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}$

Ta có đpcm.

Bài 8:

a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)

b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)

14 tháng 10 2023

a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)

\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)

\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)

\(=\dfrac{16}{3}+\dfrac{4}{3}\)

\(=\dfrac{16+4}{3}\)

\(=\dfrac{20}{3}\)

b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)

\(=\dfrac{9}{4}+\dfrac{8}{4}\)

\(=\dfrac{17}{4}\) 

c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)

\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)

\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)

\(=-\dfrac{1}{10}\) 

d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{9}{10}+\dfrac{1}{6}\)

\(=-\dfrac{11}{15}\) 

e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)

\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)

\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)

\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)

\(=3^{7-6}\)

\(=3\)

14 tháng 10 2023

\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)

\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)

\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)

`e,` Không hiểu đề á c: )

2/3A=2/3-(2/3)^2+...+(2/3)^2019-(2/3)^2020

=>5/3A=1-(2/3)^2020

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