314 Вправа 314Алгебра 8 клас Бевз 2025 · § 7 · стор. 66

314. Доведи, що значення виразу не залежить від значень змінних, які до нього входять.

а) ${\displaystyle (\frac{4}{x}-\frac{x}{4})\cdot (\frac{1}{4-x}-\frac{1}{4+x})}$;

1) ${\displaystyle \frac{4^{\mathrm{\setminus }4}}{x}-\frac{x^{\mathrm{\setminus }x}}{4}=}$${\displaystyle \frac{16-x^2}{4x}=}$${\displaystyle \frac{(x+4)(4-x)}{4x}}$;

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

=${\displaystyle \frac{x+4-4+x}{(4-x)(x+4)}=}$${\displaystyle \frac{2x}{(4-x)(x+4)}=}$${\displaystyle \frac{2x}{16-x^2}}$;

3) ${\displaystyle \frac{(x+4)(4-x)}{4x}\cdot \frac{2x}{(x+4)(4-x)}=}$${\displaystyle \frac{(x+4)(4-x)\cdot 2x}{4x\cdot (x+4)(4-x)}}$=

=${\displaystyle \frac{\cancel{2}\cancel{x}\cancel{\left(x+4\right)}\cancel{\left(4-x\right)}}{2\cdot \cancel{2}\cancel{x}\cancel{\left(x+4\right)}\cancel{\left(4-x\right)}}=}$${\displaystyle \frac{1}{2}}$;

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

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

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

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

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

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

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

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

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

4) ${\displaystyle \frac{b}{b-a}+\frac{b}{a-b}=}$${\displaystyle \frac{b}{b-a}-\frac{b}{b-a}=}$${\displaystyle \frac{b-b}{b-a}=}$${\displaystyle \frac{0}{b-a}=}$${\displaystyle 0}$;

в) ${\displaystyle (\frac{2}{3a}-\frac{3a}{2})\cdot (\frac{1}{3a-2}+\frac{1}{3a+2})}$;

1) ${\displaystyle \frac{2^{\mathrm{\setminus }2}}{3a}-\frac{{3a}^{\mathrm{\setminus }3a}}{2}=}$${\displaystyle \frac{4-9a^2}{6a}=}$${\displaystyle \frac{(3a+2)(2-3a)}{6a}}$;

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

=${\displaystyle \frac{6a}{(3a-2)(3a+2)}=}$${\displaystyle \frac{6a}{9a^2-4}}$;

3) ${\displaystyle \frac{(3a+2)(2-3a)}{6a}\cdot \frac{6a}{(3a-2)(3a+2)}=}$${\displaystyle \frac{(3a+2)(2-3a)\cdot 6a}{6a\cdot (3a-2)(3a+2)}}$=

=${\displaystyle -\frac{\cancel{6}\cancel{a}\cancel{\left(3a+2\right)}\cancel{\left(3a-2\right)}}{\cancel{6}\cancel{a}\cancel{\left(3a+2\right)}\cancel{\left(3a-2\right)}}=}$${\displaystyle -1}$;

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

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

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

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

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

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

=${\displaystyle \frac{\cancel{a}\cancel{\left(a+b\right)}}{\cancel{a}(a+2b)\cancel{\left(a+b\right)}}=}$${\displaystyle \frac{1}{a+2b}}$;

4) ${\displaystyle \frac{1}{2b+a}-\frac{1}{a+2b}=}$${\displaystyle \frac{1}{2b+a}-\frac{1}{2b+a}=}$${\displaystyle \frac{1-1}{2b+a}}$=

=${\displaystyle \frac{0}{2b+a}=}$${\displaystyle 0}$;

значення виразів сталі, тобто не залежать від змінних.