315 Вправа 315Алгебра 8 клас Бевз 2025 · § 7 · стор. 66

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

а) ${\displaystyle (\frac{2a}{b}-\frac{8b}{a})\cdot (\frac{a}{a-2b}-\frac{a}{a+2b})}$;

1) ${\displaystyle \frac{{2a}^{\mathrm{\setminus }a}}{b}-\frac{{8b}^{\mathrm{\setminus }b}}{a}=}$${\displaystyle \frac{2a^2-8b^2}{ab}=}$${\displaystyle \frac{2(a-2b)(a+2b)}{ab}}$;

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

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

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

=${\displaystyle \frac{2(a-2b)(a+2b)\cdot 4ab}{ab\cdot (a-2b)(a+2b)}=}$${\displaystyle \frac{8\cancel{a}\cancel{b}\cancel{\left(a+2b\right)}\cancel{\left(a-2b\right)}}{\cancel{a}\cancel{b}\cancel{\left(a+2b\right)}\cancel{\left(a-2b\right)}}=}$${\displaystyle 8}$;

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

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

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

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

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

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

в) ${\displaystyle (\frac{c-x}{c^2+cx}-\frac{c}{cx+x^2}):(\frac{x^2}{c^3-c{x}^2}+\frac{1}{c+x})-\frac{x-c}{x}}$;

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

=${\displaystyle \frac{x(c-x)-c^2}{cx(c+x)}=}$${\displaystyle \frac{cx-x^2-c^2}{cx(c+x)}}$=

=${\displaystyle \frac{cx-c^2-x^2}{cx(c+x)}=}$${\displaystyle -\frac{c^2-cx+x^2}{cx(c+x)}}$;

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

=${\displaystyle \frac{-x^2+c(x-c)}{c(x-c)(x+c)}=}$${\displaystyle \frac{-x^2+cx-c^2}{c(x-c)(x+c)}=}$${\displaystyle \frac{xc-x^2-c^2}{c(x-c)(x+c)}}$=

=${\displaystyle \frac{x^2-xc+c^2}{c(x+c)(c-x)}=}$${\displaystyle -\frac{x^2-xc+c^2}{c(x^2-c^2)}}$;

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

=${\displaystyle \frac{-(c^2-cx+x^2)}{cx(c+x)}\cdot \frac{c(c-x)(c+x)}{c^2-cx+x^2}}$=

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

4) ${\displaystyle \frac{x-c}{x}-\frac{x-c}{x}=}$${\displaystyle \frac{x-c-(x-c)}{x}=}$${\displaystyle \frac{x-c-x+c}{x}}$=

=${\displaystyle \frac{0}{x}=}$${\displaystyle 0}$;

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

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

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

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

3) ${\displaystyle \frac{1}{a-b}+\frac{1}{b-a}=}$${\displaystyle \frac{1}{a-b}-\frac{1}{a-b}=}$${\displaystyle \frac{1-1}{a-b}=}$${\displaystyle \frac{0}{a-b}=}$${\displaystyle 0}$;

значення виразів сталі, тобто не залежать від змінних.