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FacTeme1
a fost răspuns

Aratati ca numarul n=3^2011+2•3^2010+3^2009+3^2008 este patrat perfect

Răspuns :

3²°¹¹+2*3²°¹°+3²°°⁹+3²°°⁸=3²°°⁸x(3³+2*3²+3+1)=3²°°⁸x(27+18+3+1)=3²°°⁸x49=(3¹°°⁴)²x7²
[tex]n = {3}^{2011} + 2 \times {3}^{2010} + {3}^{2009} + {3}^{2008} [/tex]

[tex]n = {3}^{2008} ( {3}^{3} + 2 \times {3}^{2} + 3 + 1)[/tex]

[tex]n = {3}^{2008} (27 + 2 \times 9 + 4)[/tex]

[tex]n = {3}^{2008} (27 + 18 + 4)[/tex]

[tex] n = {3}^{2008} (45 + 4)[/tex]

[tex] n= {3}^{2008} \times 49[/tex]

[tex] n= {3}^{1004 \times 2} \times {7}^{2} [/tex]

[tex]n = { ({3}^{1004} )}^{2} \times {7}^{2} [/tex]

[tex]n = {( {3}^{1004} \times 7)}^{2} = > p.p[/tex]