330 Вправа 330Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

330. Спрости вираз.

а) ${\displaystyle (\frac{x^2}{x+a}-\frac{x^3}{x^2+2xa+a^2}):(\frac{x}{x+a}-\frac{x^2}{x^2-a^2})+\frac{2x^2}{x+a}}$;

1) ${\displaystyle \frac{x^2}{x+a}-\frac{x^3}{x^2+2xa+a^2}=}$${\displaystyle \frac{{x^2}^{\mathrm{\setminus }x+a}}{x+a}-\frac{x^3}{{(x+a)}^2}}$=

=${\displaystyle \frac{x^2(x+a)-x^3}{{(x+a)}^2}=}$${\displaystyle \frac{a{x}^2+x^3-x^3}{{(x+a)}^2}=}$${\displaystyle \frac{x^{2}a}{{(x+a)}^2}}$;

2) ${\displaystyle \frac{x}{x+a}-\frac{x^2}{x^2-a^2}=}$${\displaystyle \frac{x^{\mathrm{\setminus }a-x}}{x+a}+\frac{x^2}{(a-x)(x+a)}}$=

=${\displaystyle \frac{x(a-x)+x^2}{(a-x)(x+a)}=}$${\displaystyle \frac{ax-x^2+x^2}{(a-x)(x+a)}=}$${\displaystyle \frac{xa}{(a-x)(x+a)}=}$${\displaystyle \frac{xa}{a^2-x^2}}$;

3) ${\displaystyle \frac{x^{2}a}{{(x+a)}^2}:\frac{xa}{(x+a)(a-x)}=}$${\displaystyle \frac{x^{2}a}{{(x+a)}^2}\cdot \frac{(x+a)(a-x)}{xa}}$=

=${\displaystyle \frac{x^{2}a\cdot (x+a)(a-x)}{{(x+a)}^2\cdot xa}=}$${\displaystyle \frac{\cancel{x}x\cancel{a}\cancel{\left(x+a\right)}(a-x)}{\cancel{x}\cancel{a}\cancel{\left(x+a\right)}(x+a)}=}$${\displaystyle \frac{x(a-x)}{x+a}}$;

4) ${\displaystyle \frac{x(a-x)}{x+a}+\frac{2x^2}{x+a}=}$${\displaystyle \frac{ax-x^2+2x^2}{x+a}=}$${\displaystyle \frac{x^2+xa}{x+a}}$=

=${\displaystyle \frac{x\cancel{\left(x+a\right)}}{\cancel{\left(x+a\right)}}=}$${\displaystyle x}$;

б) ${\displaystyle \frac{c+2}{1-2c}-(\frac{2c}{3c+1}+\frac{3c}{1-2c})\cdot \frac{9c^2+6c+1}{5c^2+5c}}$;

1) ${\displaystyle \frac{{2c}^{\mathrm{\setminus }1-2c}}{3c+1}+\frac{{3c}^{\mathrm{\setminus }3c+1}}{1-2c}=}$${\displaystyle \frac{2c(1-2c)+3c(3c+1)}{(1-2c)(3c+1)}}$=

=${\displaystyle \frac{2c-4c^2+9c^2+3c}{(1-2c)(3c+1)}=}$${\displaystyle \frac{5c^2+5c}{(1-2c)(3c+1)}=}$${\displaystyle \frac{5c(c+1)}{(1-2c)(3c+1)}}$;

2) ${\displaystyle \frac{5c(c+1)}{(3c+1)(1-2c)}\cdot \frac{9c^2+6c+1}{5c^2+5c}}$=

=${\displaystyle \frac{5c(c+1)\cdot (9c^2+6c+1)}{(3c+1)(1-2c)\cdot (5c^2+5c)}}$=

=${\displaystyle \frac{\cancel{5}\cancel{c}\cancel{\left(3c+1\right)}(3c+1)\cancel{\left(c+1\right)}}{\cancel{5}\cancel{c}(1-2c)\cancel{\left(3c+1\right)}\cancel{\left(c+1\right)}}=}$${\displaystyle \frac{3c+1}{1-2c}}$;

3) ${\displaystyle \frac{c+2}{1-2c}-\frac{3c+1}{1-2c}=}$${\displaystyle \frac{c+2-(3c+1)}{1-2c}=}$${\displaystyle \frac{c+2-3c-1}{1-2c}}$=

=${\displaystyle \frac{\cancel{\left(1-2c\right)}}{\cancel{\left(1-2c\right)}}=}$${\displaystyle 1}$.