Proszę o zrobienie tego zadania z góry dzięki :).... Question from @pawel007007

a)
$(x + 3y) (x-3y) + (x + 2y) ^2 - (2x^2 - 5y^2) = \\ =x^2-3xy+3xy-9y^2 + x^2 + 4xy + 4y^2 - 2x^2 + 5y^2 = 4xy = \\ dla\ x = \sqrt{8}\ i\ y=\sqrt{2} \\ = 4 \cdot \sqrt{8} \cdot \sqrt{2} = 4\cdot \sqrt{16} = 2 \cdot 4 = 8$

b) $(y-3x) (y+3x) - 2x(x-y) + (y - 3)^2 +6y\\ y^2 - 9x^2 - 2x^2 + 2xy + y^2 -6y + 9 + 6y =\\= 2y^2 - 11x^2 + 2xy + 9 = \\ dla\ x = \sqrt{2}\ i\ y =\sqrt{8}\\ = 2(\sqrt{8})^2 - 11(\sqrt{2})^2 + 2\sqrt{2} \cdot \sqrt{8} + 9 = 2 \cdot 8 - 11 \cdot 2 + 2\sqrt{16} + 9 = \\ =16 - 22 + 2\cdot 4 + 9 = =16 - 22 + 8 + 9 = 11$

c) $(x+2y)(x-2y) - 4x(x-y) + (x-2y)^2 + 4xy = \\ = x^2 - 4y^2 - 4x^2+4xy + x^2 - 4xy + 4y^2 + 4xy = \\ = -2x^2 - 4y^2 +4xy = \\ dla\ x = 0,5\ i\ y= -0,5\\ = -2(0,5)^2 - 4(0,5)^2 +4(0,5) \cdot (0,5) = \\=-2 \cdot0,25 - 4\cdot 0,25 +4 \cdot 0,25 =\\ =-0,5 - 1 +1 = -0,5$

d) $(2xy + 5)^2 - (4+3xy)(3xy-4)+5x^2y^2\\ 4x^2y^2 + 20xy + 25 - (3xy + 4)(3xy-4) + 5x^2y^2 = \\ 4x^2y^2 + 20xy + 25 - (9x^2y^2 -16) + 5x^2y^2 =\\ 4x^2y^2 + 20xy + 25 - 9x^2y^2 +16 + 5x^2y^2 = \\ =20xy + 25 + 16 = 20xy + 41 = \\ dla\ x=1\frac {1}{3} = \frac{4}{3}\ i\ y= -5\frac{1}{4} = -\frac{21}{4} \\ = 20 \cdot \frac{4}{3} \cdot (-\frac{21}{4}) + 41 = 20 \cdot \frac{4}{3} \cdot (-\frac{21}{4}) + 41=\\ = 20 \cdot (-7) + 41 = -140 + 41 = -99$