352 Вправа 352Алгебра 8 клас Бевз 2025 · § 7 · стор. 70

352. Знайди значення виразу.

а) ${\displaystyle (\frac{x}{y}+\frac{y}{x}-2):(\frac{1}{y}-\frac{1}{x})}$, якщо x = 3,7, y = −1,3;

1) ${\displaystyle \frac{x^{\mathrm{\setminus }x}}{y}+\frac{y^{\mathrm{\setminus }y}}{x}-\frac{2^{\mathrm{\setminus }xy}}{1}=}$${\displaystyle \frac{x^2+y^2-2xy}{xy}=}$${\displaystyle \frac{{(x-y)}^2}{xy}}$;

2) ${\displaystyle \frac{1^{\mathrm{\setminus }x}}{y}-\frac{1^{\mathrm{\setminus }y}}{x}=}$${\displaystyle \frac{x-y}{yx}}$;

3) ${\displaystyle \frac{{(x-y)}^2}{xy}:\frac{x-y}{xy}=}$${\displaystyle \frac{{(x-y)}^2}{xy}\cdot \frac{xy}{x-y}=}$${\displaystyle \frac{{(x-y)}^2\cdot xy}{xy\cdot (x-y)}}$=

=${\displaystyle \frac{\cancel{x}\cancel{y}\cancel{\left(x-y\right)}(x-y)}{\cancel{x}\cancel{y}\cancel{\left(x-y\right)}}=}$${\displaystyle x-y}$;

якщо x = 3,7, y = −1,3, то ${\displaystyle 3,7-(-1,3)}$=

=${\displaystyle 5}$;

б) ${\displaystyle (\frac{a^2}{b^2}+\frac{b}{a}):(\frac{1}{a}-\frac{1}{b}+\frac{a}{b^2})}$, якщо a = 12, b = 8;

1) ${\displaystyle \frac{{a^2}^{\mathrm{\setminus }a}}{b^2}+\frac{b^{\mathrm{\setminus }{b}^2}}{a}=}$${\displaystyle \frac{a^3+b^3}{a{b}^2}=}$${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}}$;

2) ${\displaystyle \frac{1^{\mathrm{\setminus }{b}^2}}{a}-\frac{1^{\mathrm{\setminus }ab}}{b}+\frac{a^{\mathrm{\setminus }a}}{b^2}=}$${\displaystyle \frac{b^2-ab+a^2}{a{b}^2}}$;

3) ${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}:\frac{a^2-ab+b^2}{a{b}^2}}$=

=${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}\cdot \frac{a{b}^2}{a^2-ab+b^2}}$=

=${\displaystyle \frac{\cancel{a}\cancel{b^2}(a+b)\cancel{\left(a^2-ab+b^2\right)}}{\cancel{a}\cancel{b^2}\cancel{\left(a^2-ab+b^2\right)}}=}$${\displaystyle a+b}$;

якщо a = 12, b = 8, то ${\displaystyle 12+8=}$${\displaystyle 20}$.