1+2+3+...+x = (1 + x) * x / 2 = 666

(1 + x) * x = 666 * 2 = 1332

1332 / x - 1 = x <=> x * (x + 1) = 1332

=> x = 36

ko

31.15.49.4-31.49.40

=15.4-40

=60-40

=20

1 ) 10 \(⋮\) n

=> n \(\in\) Ư ( 10 )

Ư ( 10 ) = { 1 , 2 , 5 , 10 }

Vậy n \(\in\) { 1 ; 2 ; 5 ; 10 }

2 ) 12 : \(⋮\) ( n - 1 )

=> n - 1 \(\in\) Ư ( 12 )

=> Ư ( 12 ) = { 1 ; 12 ; 2 ; 6 ; 3 ; 4 }

Vậy n \(\in\) { 2 , 13 , 3 , 7 , 4 , 5 }

3 ) 20 \(⋮\) ( 2n + 1 )

=> 2n + 1 \(\in\) Ư ( 20 )

=> Ư ( 20 ) = { 1 ; 20 ; 2 ; 10 ; 4 ; 5 }

Các trường hợp loại , vì n \(\in\) N

Vậy n thuộc { 0 , 2 }

k minh minh giai cho

gIẢI ĐI

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

mất thời gian vậy

Dài zậy trời

a:Sửa đề: \(10^{n}+18n-1\) chia hết cho 27

Đặt \(A=10^{n}+18n-1\)

\(=\left(10^{n}-1\right)+18n=99\ldots9+18n\) (n chữ số 9)

=9(11...1+2n)⋮9

11..1+2n=n+2n=3n⋮3

=>A⋮9*3

=>A⋮27

b: Sửa đề: \(10^{n}+72n-1\)

Đặt \(B=10^{n}+72n-1\)

\(=\left(10^{n}-1\right)+72n\)

=99...9+72n(n chữ 9)

=9(11...1+8n)

11...1+8n=n+8n=9n⋮9

=>B⋮9*9

=>B⋮81

1+1/A+1/a²+1/a³+1+.../aⁿ+1

=1(1/A/a²/a³/...aⁿ)

=1.(1/a^1+2+3+...+n)

=1.(1/a^6+...+n)

=a^6+...+n

a) =43/12.1/2+1/24 b) =3/4-3/2+1/2.12/5

=43/24+1/24 =-3/4+6/5

=11/6 =9/20