171 Вправа 171Алгебра 8 клас Бевз 2025 · § 4 · стор. 38

171. Спрости вираз.

а) ${\displaystyle \frac{3}{a}-\frac{15}{a^2+5a}=}$${\displaystyle \frac{3^{\mathrm{\setminus }a+5}}{a}-\frac{15}{a(a+5)}=}$${\displaystyle \frac{3(a+5)-15}{a(a+5)}}$=

=${\displaystyle \frac{3a+15-15}{a(a+5)}=}$${\displaystyle \frac{3\cancel{a}}{(a+5)\cancel{a}}=}$${\displaystyle \frac{3}{a+5}}$;

б) ${\displaystyle \frac{3a-8}{4a-8}-\frac{a-3}{2a-4}=}$${\displaystyle \frac{3a-8}{4(a-2)}-\frac{{a-3}^{\mathrm{\setminus }2}}{2(a-2)}}$=

=${\displaystyle \frac{3a-8-2(a-3)}{4(a-2)}=}$${\displaystyle \frac{3a-8-2a+6}{4(a-2)}=}$${\displaystyle \frac{\cancel{\left(a-2\right)}}{4\cancel{\left(a-2\right)}}=}$${\displaystyle \frac{1}{4}}$;

в) ${\displaystyle \frac{16x}{4xy+3y^2}-\frac{9y}{4x^2+3xy}=}$${\displaystyle \frac{{16x}^{\mathrm{\setminus }x}}{y(4x+3y)}-\frac{{9y}^{\mathrm{\setminus }y}}{x(4x+3y)}}$=

=${\displaystyle \frac{16x^2-9y^2}{xy(4x+3y)}=}$${\displaystyle \frac{(4x-3y)\cancel{\left(4x+3y\right)}}{xy\cancel{\left(4x+3y\right)}}=}$${\displaystyle \frac{4x-3y}{xy}}$;

г) ${\displaystyle \frac{x^2-8xy}{xy-4y^2}-\frac{16y}{4y-x}=}$${\displaystyle \frac{x^2-8xy}{y(x-4y)}+\frac{{16y}^{\mathrm{\setminus }y}}{x-4y}}$=

=${\displaystyle \frac{x^2-8xy+16y^2}{y(x-4y)}=}$${\displaystyle \frac{(x-4y)\cancel{\left(x-4y\right)}}{y\cancel{\left(x-4y\right)}}=}$${\displaystyle \frac{x-4y}{y}}$.