№ 7.27 Алгебра = № 12.27 Математика
Спростіть вираз:
1. $\left(\frac{8x^2+2x}{8x^3-1}-\frac{2x+1}{4x^2+2x+1}\right)\cdot$
$\cdot\left(1+\frac{2x+1}{2x}-\frac{4x^2+10x}{4x^2+2x}\right);$
2. $\frac{p^2-2p+1}{4}\cdot$
$\cdot\left(\frac{2p}{p^3+1}:\frac{1-p}{p^2-p+1}+\frac{2}{p-1}\right):\frac{p-1}{p+1}.$
Розв'язок:
1. $\left(\frac{8x^2+2x}{8x^3-1}-\frac{2x+1}{4x^2+2x+1}\right)\cdot$
$\cdot\left(1+\frac{2x+1}{2x}-\frac{4x^2+10x}{4x^2+2x}\right)=$
$= \frac{8x^2+2x-\left(4x^2-1\right)}{\left(2x-1\right)\left(4x^2+2x+1\right)}\cdot$
$\cdot\frac{4x^2+2x+\left(2x+1\right)^2-4x^2-10x}{2x\left(2x+1\right)}=$
$= \frac{4x^2+2x+1}{\left(2x-1\right)\left(4x^2+2x+1\right)}\cdot$
$\cdot\frac{4x^2+2x+4x^2+4x+1-4x^2-10x}{2x(2x+1)}=$
$= \frac{1}{2x-1}\cdot\frac{4x^2-4x+1}{2x(2x+1)}=$
$= \frac{\left(2x-1\right)^2}{\left(2x-1\right)\cdot2x(2x+1)}=\frac{2x-1}{2x(2x+1)}.$
2. а) $\frac{2p}{p^3+1}:\frac{1-p}{p^2-p+1}=$
$= \frac{2p(p^2-p+1)}{\left(p+1\right)(p^2-p+1)\left(1-p\right)}=$
$= \frac{2p}{\left(p+1\right)\left(1-p\right)};$
б) $\frac{2p}{\left(p+1\right)\left(1-p\right)}+\frac{2}{\left(p-1\right)}=$
$= \frac{2p}{\left(p+1\right)\left(1-p\right)}-\frac{2}{\left(1-p\right)}=$
$= \frac{2p-2\left(p+1\right)}{\left(p+1\right)\left(1-p\right)}=$
$= \frac{-2}{\left(p+1\right)\left(1-p\right)};$
в) $\frac{p^2-2p+1}{4}\cdot\frac{-2}{\left(p+1\right)\left(1-p\right)}:\frac{p-1}{p+1}=$
$= \frac{\left(p-1\right)^2\left(-2\right)\left(p+1\right)}{-4\cdot\left(p+1\right)\left(p-1\right)\left(p-1\right)}=\frac{1}{2}.$