655 Вправа 655Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 194

655. Спростіть вираз:

1) ${\displaystyle \frac{{\sqrt{a}+3}^{\mathrm{\setminus }\left(\sqrt{a}+3\right)}}{\sqrt{a}-3}-\frac{{\sqrt{a}-3}^{\mathrm{\setminus }\left(\sqrt{a}-3\right)}}{\sqrt{a}+3}=}$${\displaystyle \frac{{(\sqrt{a}+3)}^2-{(\sqrt{a}-3)}^2}{{\left(\sqrt{a}\right)}^2-3^2}=}$${\displaystyle \frac{(a+6\sqrt{a}+9)-(a-6\sqrt{a}+9)}{a-9}=}$${\displaystyle \frac{a+6\sqrt{a}+9-a+6\sqrt{a}-9}{a-9}=}$${\displaystyle \frac{12\sqrt{a}}{a-9}}$;

2) ${\displaystyle \frac{\sqrt{m}+\sqrt{p}}{\sqrt{m}-\sqrt{p}}+\frac{\sqrt{m}-\sqrt{p}}{\sqrt{p}+\sqrt{m}}=}$${\displaystyle \frac{{\sqrt{m}+\sqrt{p}}^{\mathrm{\setminus }\left(\sqrt{m}+\sqrt{p}\right)}}{\sqrt{m}-\sqrt{p}}+\frac{{\sqrt{m}-\sqrt{p}}^{\mathrm{\setminus }\left(\sqrt{m}-\sqrt{p}\right)}}{\sqrt{m}+\sqrt{p}}=}$${\displaystyle \frac{{(\sqrt{m}+\sqrt{p})}^2+{(\sqrt{m}-\sqrt{p})}^2}{m-p}=}$${\displaystyle \frac{(m+2\sqrt{mp}+p)+(m-2\sqrt{mp}+p)}{m-p}=}$${\displaystyle \frac{m+2\sqrt{mp}+p+m-2\sqrt{mp}+p}{m-p}=}$${\displaystyle \frac{2m+2p}{m-p}}$;

3) ${\displaystyle \frac{\sqrt{x}+1}{1-\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}+1}=}$${\displaystyle \frac{-(\sqrt{x}+1)}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}=}$${\displaystyle \frac{-{(\sqrt{x}+1)}^2+{(\sqrt{x}-1)}^2}{(\sqrt{x}-1)(\sqrt{x}+1)}=}$${\displaystyle \frac{-(x+2\sqrt{x}+1)+(x-2\sqrt{x}+1)}{x-1}=}$${\displaystyle \frac{-x-2\sqrt{x}-1+x-2\sqrt{x}+1}{x-1}=}$${\displaystyle \frac{-4\sqrt{x}}{x-1}}$;

4) ${\displaystyle \frac{\sqrt{m}+2\sqrt{n}}{\sqrt{m}-2\sqrt{n}}-\frac{m+4n}{(\sqrt{m}-2\sqrt{n})(\sqrt{m}+2\sqrt{n})}=}$${\displaystyle \frac{{(\sqrt{m}+2\sqrt{n})}^2-(m+4n)}{m-4n}=}$${\displaystyle \frac{m+4\sqrt{mn}+4n-(m+4n)}{m-4n}=}$${\displaystyle \frac{m+4\sqrt{mn}+4n-m-4n}{m-4n}=}$${\displaystyle \frac{4\sqrt{mn}}{m-4n}}$.