Proszę o pomocDam naj ♥️♥️♥️7. Oblicz wartość wyrażenia:a) (2x + 3y) (2x – 3y) - (2x – 3y)2 dla x = .... Question from @lahcim2305

June 2022 1 35 Report

Proszę o pomoc
Dam naj ♥️♥️♥️
7. Oblicz wartość wyrażenia:
a) (2x + 3y) (2x – 3y) - (2x – 3y)2 dla x = ✓ V10 – 3, y= V10 +3,
b) (V3x - y)2 – (x - V3y)2 dla x = 16+ V2, y=
V6+ V2, y= V6 - V2.
c) (V2x - V3y)2 – (V2x – V5y)(72x + v5y) dla x = V5 - 4, y=V6,
d) (x2 – 4y2)2 – (4y2 – x2)2 dla x= v2-1, y = 2+1

Roma

Zad. 7

$(2x+3y)(2x-3y) - (2x-3y)^2= (2x-3y)(2x+3y-2x+3y) = \\ = (2x-3y) \cdot 6y =12xy - 18y^2 \\\\ x = \sqrt{\sqrt{10} -3}, \ y = \sqrt{\sqrt{10}+3} \\\\ 12xy - 18y^2 = 12\cdot \sqrt{\sqrt{10} -3} \cdot \sqrt{\sqrt{10}+3} - 18 \cdot (\sqrt{\sqrt{10}+3})^2 = \\ = 12\cdot \sqrt{(\sqrt{10} -3) \cdot (\sqrt{10}+3)} - 18 \cdot (\sqrt{10}+3)=12\cdot \sqrt{10- 9} \\ -18\sqrt{10}-54= 12-18\sqrt{10}-54=-18\sqrt{10}-42$

$(\sqrt{3}x-y)^2-(x-\sqrt{3}y)^2=3x^2-2\sqrt{3}xy + y^2 -(x^2-2\sqrt{3}xy+3y^2)= \\ =3x^2-2\sqrt{3}xy + y^2 -x^2+2\sqrt{3}xy-3y^2=2x^2-2y^2= 2 \cdot (x^2-y^2) \\\\ x =\sqrt{6} +\sqrt{2}, \ y = \sqrt{6}-\sqrt{2} \\\\ 2 \cdot (x^2-y^2) = 2 \cdot [(\sqrt{6} +\sqrt{2})^2-(\sqrt{6} -\sqrt{2})^2]=2 \cdot [6+2\sqrt{12}+2-(6-2\sqrt{12}+2)]= \\=2 \cdot (6+2\sqrt{12}+2-6+2\sqrt{12}-2)=2 \cdot 4 \sqrt{12}=8\sqrt{4 \cdot 3} = 16\sqrt{3}$

$(\sqrt{2}x-\sqrt{3}y)^2 - (\sqrt{2}x-\sqrt{5}y)(\sqrt{2}x+\sqrt{5}y)=2x^2-2\sqrt{6}xy +3y^2-(2x^2-5y^2) = \\ = 2x^2-2\sqrt{6}xy+3y^2-2x^2+5y^2=8y^2-2\sqrt{6}xy \\\\ x =\sqrt{5} - 4, \ y = \sqrt{6} \\\\8y^2-2\sqrt{6}xy= 8 \cdot (\sqrt{6})^2-2\sqrt{6} \cdot (\sqrt{5} - 4) \cdot \sqrt{6}= 8 \cdot 6-12\cdot (\sqrt{5} - 4) = \\ = 48 -12\sqrt{5} +48=96 -12\sqrt{5}= 12 \cdot (8 - \sqrt{5})$