№ 6.15 Алгебра = № 11.15 Математика
Виконайте ділення:
1. $\frac{9+6a+4a^2}{2a-1}:\frac{27-8a^3}{1-4a^2};$
2. $\frac{8+x^3}{16-x^4}:\frac{x^2-2x+4}{x^2+4};$
3. $\left(25x^2-10xy+y^2\right)∶\frac{y^2-5xy}{7};$
4. $\frac{\left(6y-4x\right)^2}{3}:\left(9y^2-12xy+4x^2\right).$
Розв'язок:
1. $\frac{9+6a+4a^2}{2a-1}:\frac{27-8a^3}{1-4a^2}=$
$= \frac{\left(9+6a+4a^2\right)\left(1-2a\right)\left(1+2a\right)}{-\left(1-2a\right)\left(3-2a\right)\left(9+6a+4a^2\right)}=$
$= -\frac{1+2a}{3-2a}=\ \frac{1+2a}{2a-3};$
2. $\frac{8+x^3}{16-x^4}:\frac{x^2-2x+4}{x^2+4}=$
$= \frac{\left(2+x\right)\left(4-{2x+x}^2\right)\left(4+x^2\right)}{\left(4-x^2\right)\left(4+x^2\right)\left(x^2-2x+4\right)}=$
$= \frac{2+x}{\left(2-x\right)\left(2+x\right)}=\frac{1}{2-x};$
3. $\left(25x^2-10xy+y^2\right)∶\frac{y^2-5xy}{7}=$
$= \frac{\left(5x-y\right)^2\cdot7}{-y\left(5x-y\right)}=$
$= \frac{7\left(5x-y\right)}{-y}=\frac{7\left(y-5x\right)}{y};$
4. $\frac{\left(6y-4x\right)^2}{3}:\left(9y^2-12xy+4x^2\right)=$
$\frac{\left((2x-3y\right))^2\cdot1}{3\cdot \left(3y-2x\right)^2}= $
$= \frac{4\cdot \left(3y-2x\right)^2}{3\cdot \left(3y-2x\right)^2}= \frac{4}{3}=1\frac{1}{3}.$