2+4+6+...+(2n)=756

=>2(1+2+3+...+n)=756

\(\Rightarrow\frac{2n\left(n+1\right)}{2}=756\)

=>n(n+1)=756

=>n=27

   2+4+6+........+2n=870

2.(2+4+6+.......+2n)=870

    2+4+6+........+2n=870:2

    2+4+6+....... +2n=435

=>n.(n+1):2=435

n.(n+1)=435.2

n.(n+1)=870

Vì 29.30=870=>n=29

\(2\left(1+2+3+...+n\right)=870\)

\(\leftrightarrow2.\frac{n\left(n-1\right)}{2}=870\)

\(\leftrightarrow n\left(n-1\right)=870=30.29\)

\(\rightarrow n=30\)

2+4+6+...+2n=870

=>2(1+2+3+...+n)=870

=>1+2+3+...+n=870:2

=>1+2+3+...+n=435

=>(n+1).n:2=435

=>(n+1).n=435.2

=>(n+1).n=870

=>(n+1).n=29.30

=>n=29

a)

\(\left(2n+1\right)^3=27\)

\(\left(2n+1\right)^3=3^3\)

\(2n+1=3\)

\(2n=3+1\)

\(2n=4\)

\(n=4\div2\)

\(n=2\)

b)

\(\left(n+2\right)^2=\left(n+2\right)^4\)

\(\left(n+2\right)^4-\left(n+2\right)^2=0\)

\(\left(n+2\right)^2\cdot\left(n+2\right)^2-\left(n+2\right)^2\cdot1=0\)

\(\left(n+2\right)^2\cdot\left[\left(n+2\right)^2-1\right]=0\)

\(\Rightarrow\left(n+2\right)^2=0hoạc\left(n+2\right)^2-1=0\)

\(\left(n+2\right)^2=0\)

\(n+2=0\)

\(n=0+2\)

\(n=2\)

\(\left(n+2\right)^2-1=0\)

\(\left(n+2\right)^2=0+1\)

\(\left(n+2\right)^2=1\)

\(n+2=1\)

\(n=1+2\)

\(n=3\)

Vậy  \(n\in\left\{2;3\right\}\)

(2n +1)³ = 3³

=> 2n + 1 = 3

sau đó dễ rùi

a) (2n + 1)^3 = 27

     (2n + 1)^3 = 3^3

     2n + 1        = 3

     2n              = 3 - 1

     2n              = 2

       n              = 2:2

       n              = 1

\(A=\left(\frac{2+2m.m}{2m}\right)=\left(\frac{2\left(1+m\right).m}{2m}\right)=1+m\)

\(B=\left(\frac{2+2n.n}{2n}\right)=\left(\frac{2\left(1.n\right).n}{2n}\right)=1.n\)

Do đó A < b => 1 + m < 1 + n => m < n

\(A=\frac{\left(2+2m\right).m}{2m}=\frac{2\left(1+m\right).m}{2m}=1+m\)

\(B=\frac{\left(2+2n\right).n}{2n}=\frac{2\left(1+n\right).n}{2n}=1+n\)

do A < b => 1 + m < 1 +n => m < n