Question

Rozwiąż równania:
a) 2[3-2(x+1)]+3[3(x-2)+1]=2+x
b)12-2(x+1)²=4(x-2)-(x-3)(2x-5)

c)x+3(2-x)/2=-1/6+-3-x/3

/-kreska ułamkowa

Answer 1

₂2[3-2(x+1)]+3[3(x-2)+1]=2+x
2(3-2-2)+3(3x-6+1)=2+x
-2+9x-18+3=2+x
7x=2+2+18-3
7x=19/:7
x=19/7



b)12-2(x+1)²=4(x-2)-(x-3)(2x-5)
12-2(x²+2x+1)=4x-8+2x-5-6x+15
12-2x²-4x-2=2
-2x²-4x+8=0
a=-2 b=-4 c=8

Δ=b²-4ac
Δ=16-(-64)
Δ=80

x₁= -b-√Δ /2a x₂=-b=√Δ /2a
x₁=4-√16x5 /-4 x₂=4+√16x5 /-4
x₁=4-4√5 /-4 x₂=4+4√5 /-4
x₁=4-√5 x₂=4+√5

Answer 2

A) 2[3-2(x+1)]+3[3(x-2)+1]=2+x
2(3-2x+2)+3(3x-6+1)=2+x
2(5-2x)+3(3x-5)=2+x
10-4x+9x-15=2+x
10-15-2=x+4x-9x
-7=-4x / ÷ (-4)
1,75=x