Question

10 și 11 rezolvare completă vă rog dau coroană​

Answer 1

 

10)  

$\displaystyle\bf\\a=4^{n+1}\cdot3^n\cdot6+12^{n+1}\cdot3+2^{2n+1}\cdot3^n\cdot4\\\\a=4^{n}\cdot4^1\cdot3^n\cdot6+12^{n}\cdot12^1\cdot3+2^{2n}\cdot2^1}\cdot3^n\cdot4\\\\a=4^{n}\cdot3^n\cdot4\cdot6+12^{n}\cdot12\cdot3+\Big(2^2\Big)^{n}\cdot2\cdot3^n\cdot4\\\\a=4^{n}\cdot3^n\cdot4\cdot6+12^{n}\cdot12\cdot3+4^{n}\cdot3^n\cdot2\cdot4\\\\a=12^{n}\cdot24+12^{n}\cdot36+12^{n}\cdot8\\\\a=12^{n}\Big(24+36+8\Big)\\\\a=68\cdot12^n\\\\a=4\cdot17\cdot12^n\\\\\boxed{\bf a~\vdots~17~oricare~ar~fi~n}$

.

11)

$\displaystyle\bf\\\textbf{Vom calcula doar paranteza a doua}\\\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot\left[\frac{3}{16}\cdot\left(-\frac{2}{3} \right)^4+\left(-\frac{1}{3} \right)^3\right]=\\\\\\=\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot\left[\frac{3}{16}\cdot\frac{16}{81} -\frac{1}{27}\right]=$

.

$\displaystyle\bf\\=\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot\left[\frac{3\cdot16}{16\cdot81} -\frac{1}{27}\right]=\\\\\\=\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot\left[\frac{3}{81} -\frac{1}{27}\right]=\\\\\\=\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot\left[\underbrace{\frac{1}{27} -\frac{1}{27}}_{=~0}\right]=\\\\\\\\=\left[\frac{5}{9}:1,\!(6)-0,\!1(6):\frac{1}{2}\right]\cdot0=\boxed{\bf0}$