Упростите выражения:
а) [tex] \frac{x+40}{x^{3}-16x } : ( \frac{x-4}{3x^{2}+11x-4 } - \frac{16}{16-x^{2} } ) [/tex]
б) [tex] \frac{y^{3}-y}{y-4}*( \frac{y-1}{2y^{2} +3y+1 } - \frac{1}{ y^{2}-1 } )[/tex]
Заранее спасибо.
а) [tex] \frac{x+40}{x^{3}-16x } : ( \frac{x-4}{3x^{2}+11x-4 } - \frac{16}{16-x^{2} } ) [/tex]
б) [tex] \frac{y^{3}-y}{y-4}*( \frac{y-1}{2y^{2} +3y+1 } - \frac{1}{ y^{2}-1 } )[/tex]
Заранее спасибо.
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Answers & Comments
1)(x-4)/[(3x-1)(x+4)] +16/[(x-4)(x+4)]=[(x-4)²-16(3x-1)]/[(3x-1)(x²-16)]=
=(x²-8x+16+48x-16)/(3x-1)(x²-16)]=(x²+40x)/(3x-1)(x²-16)]=
=x(x+40)/(3x-1)(x²-16)]
2)(x+40)/[x(x²-16)] : x(x+40)/(3x-1)(x²-16)] =
=(x+40)/[x(x²-16)]*(3x-1)(x²-16)/[x(x+40)]=(3x-1)/x²
b
1)(y-1)/[(2y+1)(y+1) -1/[(y-1)(y+1)]=[(y-1)²-1(2y+1)]/[(2y+1)(y²-1)]=
=(y²-2y+1-2y-1)/[(2y+1)(y²-1)]=(y²-4y)/[(2y+1)(y²-1)]=y(y-4)/[(2y+1)(y²-1)]
2)y(y²-1)/(y-4) *y(y-4)/[(2y+1)(y²-1)]=y²/(2y+1)