341 Вправа 341Алгебра 8 клас Бевз 2025 · § 7 · стор. 69

341. Доведи тотожність.

а) ${\displaystyle (\frac{4ab}{a-b}+a-b):(1+\frac{2b}{a-b})=}$${\displaystyle a+b}$;

1) ${\displaystyle \frac{4ab}{a-b}+\frac{{a-b}^{\mathrm{\setminus }a-b}}{1}=}$${\displaystyle \frac{4ab+(a-b)(a-b)}{a-b}}$=

=${\displaystyle \frac{4ab+a^2-2ab+b^2}{a-b}=}$${\displaystyle \frac{a^2+2ab+b^2}{a-b}=}$${\displaystyle \frac{{(a+b)}^2}{a-b}}$;

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

=${\displaystyle \frac{-b-a}{b-a}=}$${\displaystyle \frac{b+a}{a-b}}$;

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

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

тотожність доведено;

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

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

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

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

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

тотожність доведено;

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

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

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

2) ${\displaystyle \frac{(x-y)(xy(x+y)-1)}{y(x+y)}\cdot \frac{x+y}{x(x-y)}}$=

=${\displaystyle \frac{\cancel{\left(x-y\right)}(xy(x+y)-1)\cancel{\left(x+y\right)}}{y\cancel{\left(x+y\right)}x\cancel{\left(x-y\right)}}=}$${\displaystyle \frac{xy(x+y)-1}{xy}}$;

3) ${\displaystyle \frac{1}{xy}+\frac{xy(x+y)-1}{xy}=}$${\displaystyle \frac{1+xy(x+y)-1}{xy}}$=

=${\displaystyle \frac{\cancel{xy}(x+y)}{\cancel{xy}}=}$${\displaystyle x+y}$;

тотожність доведено.