ВПР 1 №23 Алгебра = ВПТ 1 №23 Математика
Спростіть вираз:
1. $\frac{16m^2}{\left(4m-1\right)\left(4m+1\right)}-$
$- \frac{8m}{16m^2-1}-\frac{1}{\left(1-4m\right)\left(1+4m\right)};$
2. $\frac{8x-9}{\left(2x+1\right)^2}-\frac{8x^3+3x-1}{\left(1+2x\right)^2}-\frac{5x-7}{1+4x^2+4x}.$
Розв'язок:
1. $\frac{16m^2}{\left(4m-1\right)\left(4m+1\right)}-$
$- \frac{8m}{16m^2-1}-\frac{1}{\left(1-4m\right)\left(1+4m\right)}=$
$= \frac{16m^2}{\left(4m-1\right)\left(4m+1\right)}-$
$- \frac{8m}{\left(4m-1\right)\left(4m+1\right)}+$
$+ \frac{1}{\left(4m-1\right)\left(4m+1\right)}=$
$= \frac{16m^2-8m+1}{\left(4m-1\right)\left(4m+1\right)}=$
$= \frac{\left(4m-1\right)^2}{\left(4m-1\right)\left(4m+1\right)}= \frac{4m-1}{4m+1};$
2. $\frac{8x-9}{\left(2x+1\right)^2}-\frac{8x^3+3x-1}{\left(1+2x\right)^2}-$
$- \frac{5x-7}{1+4x^2+4x}=\frac{8x-9}{\left(2x+1\right)^2}-$
$- \frac{8x^3+3x-1}{\left(2x+1\right)^2}-\frac{5x-7}{\left(2x+1\right)^2}=$
$= \frac{8x-9-8x^3-3x+1-5x+7}{\left(2x+1\right)^2}=$
$= \frac{-8x^3-1}{\left(2x+1\right)^2}= -\frac{8x^3+1}{\left(2x+1\right)^2}=$
$= -\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{\left(2x+1\right)^2}=$
$= \frac{2x-4x^2-1}{2x+1}.$