Question

Urgent!Va rooooog!PLEASEEEEEE!​

Answer 1

Răspuns:

Explicație:

$\bf \sqrt{ \big2^{2n} \cdot \big9^{n+1}+ \big4^{n+2}\cdot \big3^{2n}} =$

$\bf \sqrt{ \big2^{2n} \cdot \Big(\big3^2\Big)^{n+1}+ \Big(\big2^2\Big)^{n+2}\cdot \big3^{2n}} =$  

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n+2}+ \big2^{2n+4}\cdot \big3^{2n}} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot\Big(\big2^{2n-2n} \cdot\big3^{2n+2-2n}+ \big2^{2n+4-2n}\cdot \big3^{2n-2n}\Big)} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot\Big(\big2^{0} \cdot\big3^{2}+ \big2^{4}\cdot \big3^{0}\Big)} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot\Big(\big1 \cdot\big3^{2}+ \big2^{4}\cdot \big1\Big)} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot\Big(\big 9+ \big1\big6 \Big)} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot \big2\big5} =$

$\bf \sqrt{ \big2^{2n} \cdot\big3^{2n} \cdot \big5^2} =$

$\bf \big2^{n} \cdot\big3^{n} \cdot \big5 =\red{\underline{\big6^{n} \cdot \big5 \in \mathbb{N}; ~~n\in \mathbb{N}}}$