Odpowiedzi

2010-03-10T11:41:37+01:00
(x-7)²=x²-14x+49
(4x-8)²=16x²-64x+64
(2x+3)²=4x²+12x+9
(36-70)²=1156
(14x-1)²=196x²-28x+1
(126-13y)²=1576-3276y+169y²
(6y-6)²=36y²-72y+36
(3z-6y)²=9z²-36zy+36y²
(4b-4a)²=16b²-32ab+16a²
(7z-8y)²=49z²-112zy+64y²
(5k-5e)²=25k²-50ke+25e²
(4x-9)²=16x²-72x+81
(5y-16x)²=25y²-160xy+256x²
(z-7y)²=z²-14zy+49y²
(100-26)²=5476
(120-166)²=2116
(14z+5y)²=196z²+140zy+25y²
(4x+7y)²=16x²-56xy+49y²
(5y-17z)²=25y²-170yz+289z²
(7z-2)²=49z²-28z+4
2010-03-10T11:49:43+01:00
^bee oznaczać do potęgi 2
(x-7)=x^-14x+49
(4x-8)=16x^-64x+64
(2x+3)=4x^+12x+9
(36-70=? na pewno tak
(14x-1)=196x^-28x+1
(126-13y)=15876-3276y+139y^
(6y-6)=36y^-12+36
(3z-6y)=9z^-36xy+36y^
(4b-4a)=16b^-16ab+16a^
(7z-8y)=49z^-56zy+64y^
(5k-5e)=25k^-50ke+25e^
(4x-9)=16x^-72x+81
(5y-16x)=25y^-160xy+256x^
(z-7y)=z^-14zy+49y^
(100-26)=10000-5200+676=5476
(120-166)=14884-3984+27556=286456
(14z+5y)=196z^+160zy+25y^
(4x-7y)=16x^-56xy+49y^
(5y-17z)=25y^-170zy+289
(7z-2)=49z^-28z+4