Question

Z poniższych równań wyznacz 'x'.

a) ax² + bc =d
b) 5x = bx + bc -2a


z góry dzięki. ;d

Answer 1

Ax² + bc =d / -bc
ax2= -bc+d /:a
x2= -bc+d:a / √
x= √-bc+d:a



5x = bx + bc -2a / :5
x= 1/5bx + 1/5 bc - 2/5a

Answer 2

A) ax²+bc=d |-bc
ax²=d-bc |/a
x²=(d-bc)/2 |√
x=√(d-bc)/2

b) 5x=bx+bc-2a |-bx
5x-bx=bc-2a |/b
5x-x=c-2a
4x=c-2a |/4
x= (c-2a)/4
x= ½(c-a)

Answer 3

A) ax² + bc = d
ax² = d - bc /:a
x²= (d - bc)/a
x= √(d - bc)/a

b) 5x = bx + bc - 2a
5x - bx = bc - 2a
x(5 - b) = bc - 2a/:(5 - b)
x = (bc - 2a)/(5 - b)

koniec :)