Rozwiązane

Proszę o pomoc w tym zadaniu
a) 2 * (x + 4) ^ 2 =
b) (3x - 6) ^ 2 =
c) (x + y) ^ 2 - (x - y) ^ 2 =
d) 3 * (2 - y) ^ 2 + 2 * (y - 5) ^ 2 =
e) (3x - 2y) ^ 2 =
f) (x - 2) ^ 2 + 2x - 3 =

Odpowiedź :

Roks1985viz

Odpowiedź:

Szczegółowe wyjaśnienie:

a) 2 * (x + 4) ^ 2 =2(+4) ^ 2=2(+4)(+4)= 2(2+4+4+16)= 2(2+8+16)= 2^ 2+16+32

b) (3x - 6) ^ 2 =(3−6)^ 2=(3−6)(3−6)= 3(3−6)−6(3−6)= 92−18−6(3−6)= 92−18−18+36=9^ 2−36+36

c) (x + y) ^ 2 - (x - y) ^ 2 =x²+2xy+y² - (x²-2xy+y²)=4xy

d) 3*(4-4y+y^2)+2*(y^2-10y+25)=12-12y+3y^2+2y^-20y+50=5y^2-32y+62

3(−+2)2=3((−+2)(−)+2(−+2))= 3(−((−)+2)+2(−+2))= 3(−(−2+2)+2(−+2))= 3(2−2+2(−+2))= 3(2−4+4)

2(−5)(−5)= 2((−5)−5(−5))= 2(2−10+25)= 2(2−10+25)= 2^ 2−20+50

e) (3x - 2y) ^ 2 =3x^ 2-2y^ 2=9x^ 2−12xy+4y2

f) (x - 2) ^ 2 + 2x - 3 =(−2)2+2−3=(−2)−2(−2)+2−3=2−2−2(−2)+2−3=2−4+4+2−3=^ 2−2+1

Basetlaviz

[tex]a) \ 2\cdot(x+4)^{2} = 2\cdot(x^{2}+8x+16) = 2x^{2}+16x + 32\\\\\\b) \ (3x-6)^{2} = 9x^{2}-36x+36\\\\\\c) \ (x+y)^{2}-(x-y)^{2} = x^{2}+2xy + y^{2} -(x^{2}-2xy + y^{2})=\\\\=x^{2}+2xy+y^{2}-x^{2}+2xy - y^{2} =4xy\\\\\\d) \ 3\cdot(2-y)^{2} + 2\cdot(y-5)^{2} = 3\cdot(4-4y + y^{2}) + 2\cdot(y^{2}-10y + 25)=\\\\=12-12y+3y^{2}+2y^{2}-20y +50 =5y^{2}-32y + 62 \\\\\\e) \ (3x-2y)^{2} = 9x^{2}-12xy + 4y^{2}\\\\\\f) \ (x-2)^{2}+2x-3 = x^{2}-4x+4+2x-3 = x^{2}-2x+1[/tex]

Wyjaśnienie

Korzystamy ze wzorów uproszczonego mnożenia:

[tex](a+b)^{2} = a^{2}+2ab + b^{2}\\\\(a-b)^{2} = a^{2}-2ab + b^{2}[/tex]

Viz Inne Pytanie