Zadanie 3
a)
[tex]\frac{3}{x-2} + \frac{x+1}{x+2} - \frac{x^{2} +4}{x^{2} -4} = \frac{3}{x-2} + \frac{x+1}{x+2} - \frac{x^{2} +4}{(x-2) *(x+2)} =\frac{3(x+2)+(x-2)*(x+1)-(x^{2}+4)}{(x-2)*(x+2)} = \frac{3x+6+x^{2} -2x-2-x^{2} -4}{x^{2} -4} = \frac{2x+0}{x^{2} -4} =\frac{2x}{x^{2} -4}[/tex]
b)
[tex]\frac{4}{x^{2} 6x} - \frac{1-x}{2x} +\frac{x-1}{x+6} =\frac{4}{x*(x+6)} - \frac{1-x}{2x} +\frac{x-1}{x+6} = \frac{8 -(x+6)*(1-x)+2x*(x-1)}{2x*(x+6)} = \frac{8-(x-x^{2} +6-6x)+2x^{2} -2x}{2x^{2} +12x} = \frac{8-(-5x-x^{2} +6)+2x^{2}-2x}{2x^{2} +12x} = \frac{8+5x+x^{2} -6+2x^{2} -2x}{2x^{2} +12x} = \frac{2+3x3x^{2} }{2x^{2} +12x} = \frac{3x^{2} +3x+2}{2x^{2} +12x}[/tex]c)
[tex]\frac{x^{2} }{4x^{2} -9} +\frac{2-x}{2x-3} -\frac{6}{3-2x} = \frac{x^{2} }{(2x-3)*(2x+3)} + \frac{2-x}{2x-3} - \frac{6}{-(2x-3)} = \frac{x^{2} }{(2x-3)*(2x+3)} + \frac{2-x}{2x-3} + \frac{6}{2x-3} = \frac{x^{2}+(2x+3)*(2-x)+6(2x+3)}{(2x-3)*(2x+3)} = \frac{x^{2}+4x-x^{2} +6-3x+12x+18}{4x^{2} -9} = \frac{-x^{2} +13x+24}{4x^{2} -9}[/tex]
d)
[tex]\frac{2x^{2} -7}{9x^{2} -6x+1} -\frac{2-3x^{2} }{3x-1}-x =\frac{2x^{2} -7}{(3x-1)^{2} }-\frac{2-3x^{2} }{3x-1}-x=\frac{2x^{2} -7-(3x-1)*(2-3x^{2} -(3x-1)^{2}* x}{(3x-1)^{2} } = \frac{2x^{2} -7-(6x9x^{3} -2+3x^{2}) - (9x^{2}-6x+1) }{(3x-1)^{2}} =\frac{2x^{2} -7-6x+9x^{3} +2-3x^{2} -(9x^{3} -6x^{2} +x)}{(3x-1)^{2}}[/tex][tex]=\frac{-x^{2} +5-6x+9x^{3} -9x^{3}+6x^{2} -x}{(3x-1)x^{2} }=\frac{5x^{2} -5-7x}{(3x-1)^{2} }[/tex]
Mam nadzieję, że jest dobrze i trochę pomogłam ^^
