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(x+3)²+x(2-x)-6=19
x²+6x+9+2x-x²-6=19
8x+9-6-19=0
8x-16=0
8x=16
x=2
ad 2)
x²-(x-2)²=16
x²-(x²-4x+4)-16=0
x²-x²+4x-4-16=0
4x-20=0
4x=20
x=5
ad 3)
(3x-1)²+5(4-x)²-(2x-5)(2x+5)=10
9x²-6x+1+5(16-8x+x²)-(4x²-5²)-10=0
9x²-6x+1+80-40x+5x²-4x²+25-10=0
10x²-46x+96=0
Δ=46²-4·10·96 = 2116 - 3840 = -1724 < 0 - brak pierwiastków
ad 4)
3x(2x-3)-5(x-1)²-(x+2)(x-2)=0
6x²-9x-5(x²-2x+1)-(x²-4)=0
6x²-9x-5x²+10x-5-x²+4=0
x-1=0
x=1
ad 5)
to jest to samo co zad 3)