1A [tex]\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}-2 }{3-2}=\sqrt{6}-2[/tex] B [tex]\frac{2\sqrt{3}+6\sqrt{2}}{2-12}=\frac{-2\sqrt{3}-6\sqrt{2} }{10}[/tex] C[tex]\frac{1+\sqrt{2} }{3+\sqrt{2}}=\frac{(1+\sqrt{2})(3-\sqrt{2})}{9-2}=\frac{1-2\sqrt{2} }{7}[/tex] 2[tex]-(4x+3)(4x-3)+8(1-\sqrt{2}x)^{2}=1.. -(16x^{2} -9)+8(1-2\sqrt{2}x+2x^{2})=1.. 16\sqrt{2}x=16..x=\frac{16}{16\sqrt{2}}=\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2}[/tex] 3[tex]-9(4-4x+x^{2} )-(1-9x^{2})\leq 11\\ .. -36+36x-9x^{2} -1+9x^{2} \leq 11.. 36x\leq 48 ..x\leq \frac{48}{36}.. x\leq \frac{4}{2}[/tex]