327 Вправа 327Алгебра 8 клас Бевз 2025 · § 7 · стор. 68

327. Спрости вираз.

а) ${\displaystyle (1-\frac{2x-3y}{2x+3y}):(\frac{2xy}{4x^2-9y^2}+\frac{y}{3y-2x})}$;

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

=${\displaystyle \frac{2x+3y-2x+3y}{2x+3y}=}$${\displaystyle \frac{6y}{2x+3y}}$;

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

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

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

3) ${\displaystyle \frac{6y}{2x+3y}:\frac{3y^2}{(2x+3y)(3y-2x)}}$=

=${\displaystyle \frac{6y}{2x+3y}\cdot \frac{(2x+3y)(3y-2x)}{3y^2}=}$${\displaystyle \frac{6y\cdot (2x+3y)(3y-2x)}{(2x+3y)\cdot 3y^2}}$=

=${\displaystyle \frac{2\cdot \cancel{3}\cancel{y}\cancel{\left(3y+2x\right)}(3y-2x)}{\cancel{3}\cancel{y}y\cancel{\left(3y+2x\right)}}=}$${\displaystyle \frac{2\left(3y-2x\right)}{y}}$;

б) ${\displaystyle (\frac{a^2+ab}{a^3+a^{2}b+a{b}^2+b^3}+\frac{b}{a^2+b^2}):(\frac{1}{a-b}-\frac{2ab}{a^3-a^{2}b+a{b}^2-b^3})}$;

${\displaystyle a^3+a^{2}b+a{b}^2+b^3=}$${\displaystyle a^2(a+b)+b^2(a+b)}$=

=${\displaystyle (a+b)(a^2+b^2)}$;

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

=${\displaystyle (a-b)(a^2+b^2)}$;

1) ${\displaystyle \frac{a^2+ab}{a^3+a^{2}b+a{b}^2+b^3}+\frac{b}{a^2+b^2}}$=

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

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

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

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

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

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

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

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