Question

URGENT va roog exercițiile din poza!!! 100 de puncte!!!

Answer 1

L-am făcut doar pe 3.

Cele 2 exerciții sunt de dificultate mare.

$a = \sqrt{3-\sqrt{8}}= \sqrt{(1-\sqrt{2})^2} =|1-\sqrt{2}| = \sqrt{2}-1\\ \\ b = \sqrt{6-\sqrt{32}}= \sqrt{(2-\sqrt{2})^2} = |2-\sqrt{2}| = 2-\sqrt{2}\\ \\ \\a^{10}+a^{9}b+a^{8}b+...+a^2b-2b+1=\\ \\ =a^{9}(a+b)+a^{8}b+...+a^2b-2b+1\\ \\ = a^9(\sqrt{2}-1+2-\sqrt{2})+a^{8}b+...+a^2b-2b+1\\ \\ =a^9+a^{8}b+...+a^2b-2b+1\\ \\ =a^8(a+b)+a^7b+...+a^2b-2b+1\\ \\ = a^8+a^7b+...+a^2b-2b+1\\\\~\,\vdots\quad\text{(Analog)}\\\\=a^2(a+b)-2b+1\\\\= a^2-2b+1\\ \\ = 3-\sqrt{8}-2(2-\sqrt{2})+1\\ \\ =3-2\sqrt{2}-4+2\sqrt{2}+1\\ \\ =\boxed{0}$

4.

$\dfrac{x^2}{y^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{x^2}= \dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}\\ \\ \Leftrightarrow\,\Big(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\Big)^2 - \dfrac{2xy}{yz}-\dfrac{2xz}{yx}-\dfrac{2yz}{zx} = \dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}\\ \\ \Leftrightarrow\,\Big(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\Big)^2 - \dfrac{2x}{z}-\dfrac{2z}{y}-\dfrac{2y}{x} = \dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}$

$\Leftrightarrow\, \Big(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\Big)^2 = 3\Big(\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{z}{y}\Big)$