357 Вправа 357Алгебра 8 клас Бевз 2025 · § 7 · стор. 71

357. Виконай підстановку і спрости.

а) ${\displaystyle \frac{x-a}{x-b}}$, де ${\displaystyle x=}$${\displaystyle \frac{ab}{a+b}}$;

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

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

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

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

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

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

б) ${\displaystyle \frac{\frac{a}{b}-x}{\frac{a}{b}+x}}$, де ${\displaystyle x=}$${\displaystyle \frac{a-b}{a+b}}$;

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

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

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

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

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

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

в) ${\displaystyle \frac{ax}{a+x}-\frac{bx}{b-x}}$, де ${\displaystyle x=}$${\displaystyle \frac{ab}{a+b}}$;

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

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

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

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

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

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

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

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

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

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

г) ${\displaystyle (\frac{x+2b}{x-2b}+\frac{x+2a}{x-2a}):\frac{x}{2}}$, де ${\displaystyle x=}$${\displaystyle \frac{4ab}{a+b}}$;

1) ${\displaystyle x+2b=}$${\displaystyle \frac{4ab}{a+b}+\frac{{2b}^{\mathrm{\setminus }a+b}}{1}=}$${\displaystyle \frac{4ab+2b(a+b)}{a+b}}$=

=${\displaystyle \frac{4ab+2ab+2b^2}{a+b}=}$${\displaystyle \frac{6ab+2b^2}{a+b}=}$${\displaystyle \frac{2b(3a+b)}{a+b}}$;

2) ${\displaystyle x-2b=}$${\displaystyle \frac{4ab}{a+b}-\frac{{2b}^{\mathrm{\setminus }a+b}}{1}=}$${\displaystyle \frac{4ab-2b(a+b)}{a+b}}$=

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

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

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

4) ${\displaystyle x+2a=}$${\displaystyle \frac{4ab}{a+b}+\frac{{2a}^{\mathrm{\setminus }a+b}}{1}=}$${\displaystyle \frac{4ab+2a(a+b)}{a+b}}$=

=${\displaystyle \frac{4ab+2a^2+2ab}{a+b}=}$${\displaystyle \frac{2a^2+6ab}{a+b}=}$${\displaystyle \frac{2a(a+3b)}{a+b}}$;

5) ${\displaystyle x-2a=}$${\displaystyle \frac{4ab}{a+b}-\frac{{2a}^{\mathrm{\setminus }a+b}}{1}=}$${\displaystyle \frac{4ab-2a(a+b)}{a+b}}$=

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

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

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

7) ${\displaystyle \frac{3a+b}{a-b}+\frac{a+3b}{b-a}=}$${\displaystyle \frac{3a+b}{a-b}-\frac{a+3b}{a-b}}$=

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

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

8) ${\displaystyle 2:\frac{x}{2}=}$${\displaystyle 2:\frac{2ab}{a+b}=}$${\displaystyle 2\cdot \frac{a+b}{2ab}=}$${\displaystyle \frac{\cancel{2}(a+b)}{\cancel{2}ab}=}$${\displaystyle \frac{a+b}{ab}}$.