Question

Calcule o valor da seguinte expressão númerica:
1-17x(6,6 - 6,5999...)^20 + [√2+1/√2-1]x(2√2-3) + [(√7-4)^0-1]

Me ajudem por favor!

Answer 1

Farei por partes para ficar mais claro.
$6,5999...= \frac{659-65}{90}= \frac{594^{:18}}{90_{:18}} = \frac{33}{5} \\ ---------------------------- \\6,6= \frac{66^{:2}}{10_{:2}} = \frac{33}{5} \\ ---------------------------- \\(6,6 - 6,5999...)^{20}=\Big( \frac{33}{5}- \frac{33}{5}\Big)^{20}=0^{20}=0$

$\\ ------------------------\\ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1} = \frac{ \sqrt{2}+1 }{ \sqrt{2}-1}\cdot \frac{ \sqrt{2}+1 }{ \sqrt{2}+1}= \frac{ (\sqrt{2})^2+2\cdot\sqrt{2}\cdot 1+1^2 }{(\sqrt{2})^2-1^2}\\ \\= \frac{2+2\sqrt{2}+1}{2-1} =\frac{3+2\sqrt{2}}{1} =3+2\sqrt{2}\\ \\ -------------------------\\ \\ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1}\times( 2\sqrt{2}-3)=(3+2\sqrt{2})\times( 2\sqrt{2}-3)=(3+2\sqrt{2})\times( -3+2\sqrt{2})\\ \\=(3+2\sqrt{2})\times( -3+2\sqrt{2})=(3+2\sqrt{2})\times(-1)\times( 3-2\sqrt{2})$
$=(3+2\sqrt{2})\times(-1)\times( 3-2\sqrt{2})=(-1)\times(3+2\sqrt{2})\times( 3-2\sqrt{2})\\ \\=(-1)\times[3^2-(2\sqrt{2})^2]=(-1)\times[9-4\cdot2]=(-1)\times[9-8]\\ \\=-1\times1=-1$

$\\------------------------\\(\sqrt{7}-4)^0=1\\ ------------------------\\ \big[(\sqrt{7}-4)^0-1\big]=[1-1]=0$

Agora é só substituir:
$\ \ \ 1-17\times(6,6-6,5999...)^{20}+ \frac{\sqrt{2}+1} {\sqrt{2}-1}\times(2\sqrt{2}-3)+\Big[(\sqrt{7}-4)^0-1\Big] \\ \\=1-17\times0+ (-1)+0 \\ \\=1-0-1+0 \\ \\=0$