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AH
Akai Haruma
Giáo viên
25 tháng 3 2023

Lời giải:
$A=\frac{1}{4}(1-3+3^2-3^3+...+3^{2022}-3^{2023})$

$3A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024})$

$3A+A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024}+1-3+3^2-3^3+...+3^{2022}-3^{2023})$

$4A=\frac{1}{4}(1-3^{2024})$

$A=\frac{1}{16}(1-3^{2024})$

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: )

13 tháng 8 2019

3.

a) \(\left(x-1\right)^3=125\)

=> \(\left(x-1\right)^3=5^3\)

=> \(x-1=5\)

=> \(x=5+1\)

=> \(x=6\)

Vậy \(x=6.\)

b) \(2^{x+2}-2^x=96\)

=> \(2^x.\left(2^2-1\right)=96\)

=> \(2^x.3=96\)

=> \(2^x=96:3\)

=> \(2^x=32\)

=> \(2^x=2^5\)

=> \(x=5\)

Vậy \(x=5.\)

c) \(\left(2x+1\right)^3=343\)

=> \(\left(2x+1\right)^3=7^3\)

=> \(2x+1=7\)

=> \(2x=7-1\)

=> \(2x=6\)

=> \(x=6:2\)

=> \(x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

13 tháng 8 2019

Giúp mk với nha các bạn

28 tháng 10 2021

\(\left\{{}\begin{matrix}\left(-\dfrac{1}{4}\right)^0=1\\-2\dfrac{1}{3^2}=-2+\dfrac{1}{9}=-\dfrac{19}{9}\\0,5^3=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\\-1\dfrac{1}{3^4}=-1+\dfrac{1}{81}=-\dfrac{80}{81}\end{matrix}\right.\)

8 tháng 9 2021

toi ko co the bt day nh vau ko dau

11 tháng 9 2020

a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)

9 tháng 10 2020

 1/2 x 2 mũ n cộng 4 x 2 mũ n = 9 x 2 mũ n