zad 3
chodzi mi o przykłady
b, c i d
w b ma wyjsc 1
w c 5
w d 0,3
w zalącznikach przykłady
z gory thx
zad. 3
b)
(x-3)² - (x-7)/2 = 7 – (x+1)(1-x) /·2
2(x²-6x+9) - x + 7 = 14 – 2(1+x)(1-x)
2x2-12x+18 – x + 7 = 14 – 2(1-x2)
2x2-13x+25 = 14 – 2 + 2x2
2x2 – 13x – 2x2 = 12 – 25
-13x = -13 /:(-13)
x= 1
c)
(y-1)² - (y+4)/3 - 4 = (y-4)(4+y) /*3
3(y2-2y+1) – y – 4 – 12 = 3(y-4)(y+4)
3y2 – 6y + 3 – y – 16 = 3(y2-16)
3y2 – 7y – 13 = 3y2 – 48
3y2 – 3y2 – 7y = -48 + 13
-7y = -35 /:(-7)
y = 5
d)
(3y-2)²/4 - (y+1)² = 1,25(y-1)(y+1) – 0,5²
(9y2 – 12y + 4)/4 – (y2 + 2y + 1) = 1,25(y2 – 1) – 0,25 / *4
9y2 – 12y + 4 – 4(y2 + 2y +1) = 5(y2 – 1) – 1
9y2 – 12y + 4 – 4y2 - 8y - 4 = 5y2 – 5 – 1
5y2 – 20y = 5y2 – 6
5y2 - 5y2 – 20y = -6
-20y = -6 /:(-20)
y = 6/20
y = 3/10
y = 0,3
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zad. 3
b)
(x-3)² - (x-7)/2 = 7 – (x+1)(1-x) /·2
2(x²-6x+9) - x + 7 = 14 – 2(1+x)(1-x)
2x2-12x+18 – x + 7 = 14 – 2(1-x2)
2x2-13x+25 = 14 – 2 + 2x2
2x2 – 13x – 2x2 = 12 – 25
-13x = -13 /:(-13)
x= 1
c)
c)
(y-1)² - (y+4)/3 - 4 = (y-4)(4+y) /*3
3(y2-2y+1) – y – 4 – 12 = 3(y-4)(y+4)
3y2 – 6y + 3 – y – 16 = 3(y2-16)
3y2 – 7y – 13 = 3y2 – 48
3y2 – 3y2 – 7y = -48 + 13
-7y = -35 /:(-7)
y = 5
d)
(3y-2)²/4 - (y+1)² = 1,25(y-1)(y+1) – 0,5²
(9y2 – 12y + 4)/4 – (y2 + 2y + 1) = 1,25(y2 – 1) – 0,25 / *4
9y2 – 12y + 4 – 4(y2 + 2y +1) = 5(y2 – 1) – 1
9y2 – 12y + 4 – 4y2 - 8y - 4 = 5y2 – 5 – 1
5y2 – 20y = 5y2 – 6
5y2 - 5y2 – 20y = -6
-20y = -6 /:(-20)
y = 6/20
y = 3/10
y = 0,3