Răspuns:
a) (x-6)²+2x(5x-1)-(4x-2)(3x+1)=x²-12x+36+10x²-2x-(12x²-2x-2)=
=11x²-14x+36-12x²+2x+2=-x²-12x+38=-x²-12x+2+4+10=-12(x-2)-(x-2)(x+2)+10=10-(x-2(12+2+2)=10-(x-2)(x+14)
b) (2x-√6)²-(2x-√6)(2x+√6)+√6(4x-1)=(2x-√6)(2x-√6-2x-√6)+√6(4x-1)=(2x-√6(-2√6)+√6(4x-1)=-4√6x+12+4√6x-√6=12-√6
c) [tex]\sqrt{3-2\sqrt{2} } +\sqrt{3+2\sqrt{2} } = \sqrt{3-\sqrt{8} } +\sqrt{3+\sqrt{8} } =\sqrt{\frac{3+\sqrt{9-8}}{2}}-\sqrt{\frac{3-\sqrt{9-8}}{2}}+ \sqrt{\frac{3+\sqrt{9-8}}{2}}+\sqrt{\frac{3-\sqrt{9-8}}{2}}=2\sqrt{\frac{3+\sqrt{9-8}}{2}}=2\sqrt{\frac{3+1}{2}}=2\sqrt{2}[/tex]
Explicație pas cu pas:
