a)
4x²+12x+9-2x²-2x+2x+2+15=12-12x+3x²-x²-2x-1-11
4x²+12x+9-2x²+2+15=12-12x+3x²-x²-2x-1-11
2x²+12+26=12-12x+3x²-x²-2x-1-11
2x²+12+26=-14x+2x²
12x+26=-14x
26=-14x-12x=-26x
-1=x
x=-1
[tex]b) \\ \\ \frac{ \sqrt{2}x + 3 }{3} - \frac{x}{ \sqrt{2} } + \frac{3x}{ \sqrt{18} } = \frac{9 - \sqrt{2}x }{6} \\ \\ \frac{ \sqrt{2}x + 3 }{3} - \frac{x}{ \sqrt{2} } + \frac{3x}{3 \sqrt{2} } = \frac{9 - \sqrt{2}x }{6} \\ \\ \frac{ \sqrt{2x + 3} }{3} - \frac{x}{ \sqrt{2} } + \frac{x}{ \sqrt{2} } = \frac{9 - \sqrt{2}x }{6} \\ \\ \frac{ \sqrt{2}x + 3 }{3} = \frac{9 - \sqrt{2}x }{6} \\ \\ 1 + \frac{ \sqrt{2x} }{3} = \frac{9 - \sqrt{2}x }{6} \\ \\ 6 + 2 \sqrt{2}x = 9 - \sqrt{2}x \\ \\ 2 \sqrt{2}x = 9 - \sqrt{2}x - 6 = - \sqrt{2}x + 3 \\ \\ 2 \sqrt{2}x + \sqrt{2}x = 3 \\ \\ 3 \sqrt{2}x = 3 \\ \\ \sqrt{2}x = 1 \\ \\ x = \frac{1}{ \sqrt{2} } [/tex]
