[tex] \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} [/tex]
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.
[tex] \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}} [/tex]
[tex] \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 [/tex]
[tex] \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} [/tex]
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5)
[tex] \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} [/tex]