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27 tháng 9 2022

=0+0+.......+0+2000

=2000

mik chỉ bt thế thui

\(B=2m^6+3m^3n^3+n^6+n^3\)

\(=2m^6+2m^3n^3+m^3n^3+n^6+n^3\)

\(=2m^3\left(m^3+n^3\right)+n^3\left(m^3+n^3\right)+n^3\)

\(=2m^3+2n^3\)

=2

27 tháng 3 2018

Ta có:\(5^n.2,5-30.5^n-6.5^n-1=5^n.\left(25-30-6\right)-1=5^n.\left(-11\right)-1\)-1

14 tháng 3 2022

giúp mình với, đi mà mọi ngườikhocroi

14 tháng 3 2022

Xx4/5=1/2

       X=1/2:4/5

       X=5/8

9 tháng 1 2022

\(\left(n-4\right)⋮\left(n-1\right)\Rightarrow\left(n-1-3\right)⋮\left(n-1\right)\)

\(Mà\left(n-1\right)⋮\left(n-1\right)\Rightarrow-3⋮\left(n-1\right)\Rightarrow n-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\Rightarrow n\in\left\{-2;0;2;4\right\}\)

 

9 tháng 1 2022

<=> n-1ϵ(1,-1,3,-3)

23 tháng 5 2020

\(a, 10^{n+1} -6.10 ^n\)

= \(10^n (10-6)=4.10^n\)

\(B/ 2^{n+3} + 2^{n+2} - 2^{n+1} +2^n\)

= \(2^n (2^3+2^2-2+1)\)

= \(2^n (8+4-2+1)\)

\(= 11.2^n\)

\(C/ 90.10^k - 10^{k +2} + 10^{k +1} \)

\(= 10^k(90-2+1)\)

= \(89.10^k\)

\(D/ 2,5 . 5^{n-3} . 10+5^n -6 .5^{n-1}\)

\(= 5.5.5^{n-3} +5^n-6.5^{n-1}\)

= \(5^2 .5^{n-3}+5^n-6.5^{n-1} \)

= \(5^{n-3+2}+5^n -6.5^{n-1}\)

\(= 5^{n-1}(1+5-6)\)

= \(5^{n-1}.0\)

= 0

24 tháng 5 2020

cảm ơn ạ

16 tháng 12 2018

\(n^2+5⋮n+1\Leftrightarrow n^2-1+6⋮n+1\)

\(\Leftrightarrow\left(n-1\right)\left(n+1\right)+6⋮n+1\)

\(\Leftrightarrow6⋮n+1\) (  vì \(\left(n-1\right)\left(n+1\right)⋮n+1\) ) 

\(\Leftrightarrow n+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Đến đây bn lập bảng rồi xét để tìm n

14 tháng 6 2023

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

14 tháng 6 2023

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

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

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)