Question

Bună! As dori puțin ajutor la exercițiile 5,6 și 7, cât de curând. Ofer 30 de puncte. Sper sa ma puteți ajuta!

Answer 1

5)

$a)\quad (2x^2-3x+1)-(-4x^2+4x+2) =\\ =2x^2-3x+1+4x^2-4x-2 =\\ =6x^2-7x-1\\ \\ b)\quad (3x-1)^2+(2x+3)^2 = \\ =9x^2-6x+1+4x^2+12x+9 = \\ =13x^2+6x+10\\ \\ c)\quad (5x^2-2)(5x^2+2)-(1-5x^2)^2 =\\ =(5x^2)^2 - 2^2 - (1-5x^2)^2 =\\=25x^4-4-(1-10x^2+25x^4) =\\=25x^4-4-1+10x^2-25x^4 =\\=10x^2-5$

6)

$a)\quad 3x^2-6x = 3x(x-2)\\ b)\quad 49-x^2 = 7^2-x^2 = (7-x)(7+x) \\\\c)\quad (3x+1)^2 - 25 = (3x+1)^2 - 5^2 = \Big[(3x+1)-5\Big]\cdot \Big[(3x+1)+5\Big] =\\ =(3x-4)(3x+6)\\\\ d)\quad x^2-7x+12 = x^2-4x-3x+12 = x(x-4)-3(x-4) =\\ = (x-4)(x-3)$

7)

$a)\quad 4\sqrt{7} - 6\sqrt{7} = -2\sqrt{7}\\\\ b)\quad 8\sqrt{3}-2\sqrt{5}+\sqrt{12}-3\sqrt{20}= 8\sqrt{3}-2\sqrt{5}+2\sqrt{3} -3\cdot 2\sqrt{5} =\\=8\sqrt{3}-2\sqrt{5}+2\sqrt{3}-6\sqrt{5} = 10\sqrt{3}-8\sqrt{5}$

$c)\quad (4\sqrt{5}+3\sqrt{3}-1)^2-2(5\sqrt{2}+3)(5\sqrt{2}-3) = \\ =(4\sqrt{5}+3\sqrt{3}-1)^2 - 2\Big[(5\sqrt{2})^2 - 3^2\Big] = \\ = (4\sqrt{5}+3\sqrt{3}-1)^2-2(50-9) = \\ =(4\sqrt{5})^2+(3\sqrt{3})^2+(-1)^2+2(12\sqrt{15}-4\sqrt{5}-3\sqrt{3}) -82 = \\= 80+27+1+24\sqrt{15}-8\sqrt{5}-6\sqrt{3} - 82 = \\ =26+24\sqrt{15}-8\sqrt{5}-6\sqrt{3}$