Rozwiąż równania: ( znak / oznacza kreskę ułamkową w równaniach)
a) 3-x/2+x=7-3x/5+x
b) 2/x-3-5/x+3=12/x2-9
a)
3-x / 2+x = 7-3x/5+x
(3-x)(5+x) = (2+x)(7-3x)
15 + 3x – 5x - x² = 14 – 6x + 7x – 3x²
15 + 3x – 5x - x² - 14 + 6x - 7x + 3x² = 0
2x² - 3x + 1 = 0
∆ = b² - 4ac
∆ = (-3)² - 4 * 2 * 1 = 9 – 8 = 1
x₁ = -b-√∆/2a
x₁ = 3 – 1/ 2*2 = 2/4 = ½
x₂ = -b+√∆/2a
x₂ = 3 + 1/ 2*2 = 4/4 = 1
b)
2/x-3 – 5/x+3 = 12/x² -9
2(x+3) – 5(x-3)/(x -3)(x+3) = 12/x² -9
2(x+3) – 5(x-3)/x² -9 = 12/x² -9 / *(x²-9)
2(x+3) – 5(x-3) = 12
2x + 6 – 5x + 15 -12 =0
-3x = 9 / : (-3)
x = -3
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a)
3-x / 2+x = 7-3x/5+x
(3-x)(5+x) = (2+x)(7-3x)
15 + 3x – 5x - x² = 14 – 6x + 7x – 3x²
15 + 3x – 5x - x² - 14 + 6x - 7x + 3x² = 0
2x² - 3x + 1 = 0
∆ = b² - 4ac
∆ = (-3)² - 4 * 2 * 1 = 9 – 8 = 1
x₁ = -b-√∆/2a
x₁ = 3 – 1/ 2*2 = 2/4 = ½
x₂ = -b+√∆/2a
x₂ = 3 + 1/ 2*2 = 4/4 = 1
b)
2/x-3 – 5/x+3 = 12/x² -9
2(x+3) – 5(x-3)/(x -3)(x+3) = 12/x² -9
2(x+3) – 5(x-3)/x² -9 = 12/x² -9 / *(x²-9)
2(x+3) – 5(x-3) = 12
2x + 6 – 5x + 15 -12 =0
-3x = 9 / : (-3)
x = -3