\(^{2n}\).(-1)\(^n\).(-1)
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24 tháng 7 2017

\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)

\(A=\left(-1\right)^{2n+n+n+1}\)

\(A=\left(-1\right)^{4n+1}\)

\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)

\(B=0\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=0\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)

\(D=1999^0\)

\(D=1\)

21 tháng 8 2020

a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)

b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)

\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)

c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)

\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)

16 tháng 9 2017

cái này mà bạn ko biết làm á, bấm máy tính tạch tạch mấy phát là ra mà

17 tháng 9 2017

lười làm nên nhờ mấy bạn giải dùm

14 tháng 10 2018

\(x^2+\left(y-\dfrac{1}{10}\right)^{2018}=0\\ \Leftrightarrow x^2+\left[\left(y-\dfrac{1}{10}\right)^{1009}\right]^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^{1009}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

23 tháng 9 2017

a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)

\(5x+1=\pm\dfrac{6}{9}\)

+) \(5x+1=\dfrac{6}{9}\)

\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{9}\)

\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)

+) \(5x+1=\dfrac{-6}{9}\)

\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)

\(5x=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)

vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)

27 tháng 11 2022

a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)

\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)

\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)

\(=\dfrac{124}{15}\)

b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)

\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)

\(=-\dfrac{71}{375}\)

c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)

\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)

=1+2/5

=7/5

d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)

e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)

25 tháng 6 2019

bạn ơi cho mk hỏi 1 bài làm giúp mk đc ko vậy ạ

25 tháng 6 2019

2n  là số chẳn , n và n+1 n chẳn thì n+1 là lẻ và ngược lại nên A = -1

15 tháng 9 2017

1/2m = 1/32

1/2m = 1/25

=> m =5

18 tháng 7 2017

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

18 tháng 7 2017

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

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

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt haha

2 tháng 2 2019

iúp mình vs help me >3

3 tháng 2 2019

mk làm rùi nên mn k cần giúp nx đâu.Hihi