Question

Va rog !
Am mare nevoie de voi toti!

Answer 1

$\displaystyle\\ 4a)\\\\ \frac{\dfrac{3x}{4}-1}{3}+\frac{x}{4} =\frac{4x-3}{6}\\\\ \frac{\dfrac{3x-4}{4}}{3}+\frac{x}{4} =\frac{4x-3}{6}\\\\ \frac{3x-4}{4\cdot3}+\frac{x}{4} =\frac{4x-3}{6}\\\\ \frac{3x-4}{12}+\frac{3x}{12} =\frac{2(4x-3)}{12}~~~\Big|\cdot 12\\\\ 3x-4+3x=8x-6\\ 3x+3x-8x=-6+4\\ -2x=-2~~\Big|\cdot (-1)\\ 2x=2\\\\ x=\frac{2}{2}\\\\ \boxed{x=1}$

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$\displaystyle\\ 4b)\\ A=\left[\Big(2+3\sqrt{2}\Big)^{2018}+\frac{1}{\Big(2-3\sqrt{2}\Big)^{2018}} \right] \cdot \frac{\Big(4-6\sqrt{2}\Big)^{2018}}{4^{1009}}\\\\ A=\frac{\Big(2+3\sqrt{2}\Big)^{2018}\cdot \Big(2-3\sqrt{2}\Big)^{2018}+1}{\Big(2-3\sqrt{2}\Big)^{2018}} \right] \cdot \frac{\Big(2(2-3\sqrt{2})\Big)^{2018}}{\Big(2^2\Big)^{1009}}$

$\displaystyle\\ A=\frac{\Big(2+3\sqrt{2}\Big)^{2018}\cdot \Big(2-3\sqrt{2}\Big)^{2018}+1}{\Big(2-3\sqrt{2}\Big)^{2018}} \right] \cdot \frac{2^{2018}\cdot\Big(2-3\sqrt{2}\Big)^{2018}}{2^{2018}}\\ \text{Simplificam.}\\ A=\frac{\Big(2+3\sqrt{2}\Big)^{2018}\cdot \Big(2-3\sqrt{2}\Big)^{2018}+1}{1}\cdot \frac{1\cdot 1}{1} \\\\ A=\Big(2+3\sqrt{2}\Big)^{2018}\cdot \Big(2-3\sqrt{2}\Big)^{2018}+1$

$\displaystyle\\ A=\left[\Big(2+3\sqrt{2}\Big)\Big(2-3\sqrt{2}\Big)\right]^{2018}+1\\\\ A=\Big(4-9\cdot2\Big)^{2018}+1\\\\ A=\Big(4-18\Big)^{2018}+1\\\\ A=\Big(-14\Big)^{2018}+1\\ \text{Exponentul 2018 este un numar par.}\\ A=\Big(14\Big)^{2018}+1\\\\ \boxed{A\in Z}$

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5)

$\displaystyle\\ \frac{\sqrt{8-2\sqrt{15}}-\sqrt{6+2\sqrt{5}}+\sqrt{4-2\sqrt{3}}}{x-2}=\\\\ =\frac{\sqrt{5+3-2\sqrt{5}\cdot\sqrt{3}}-\sqrt{5+1+2\sqrt{5}\cdot1}+\sqrt{3+1-2\sqrt{3}\cdot1}}{x-2}=\\\\ =\frac{\sqrt{\Big(\sqrt{5}-\sqrt{3}\Big)^2}-\sqrt{\Big(\sqrt{5}+1\Big)^2}+\sqrt{\Big(\sqrt{3}-1\Big)^2}}{x-2}=\\\\ =\frac{\sqrt{5}-\sqrt{3}-\Big(\sqrt{5}+1\Big)+\sqrt{3}-1}{x-2}=\\\\ =\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-1+\sqrt{3}-1}{x-2}=\\\\ =\frac{-1 -1}{x-2}=\frac{-2}{x-2}\\\\ \frac{-2}{x-2}\in Z\\\\ \boxed{x=0}$