635° Вправа 635Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 190

635. Спростіть вираз:

1) ${\displaystyle \frac{\sqrt{mn}-n}{\sqrt{m}-\sqrt{n}}=}$${\displaystyle \frac{\sqrt{n}\cdot \sqrt{m}-\sqrt{n}\cdot \sqrt{n}}{\sqrt{m}-\sqrt{n}}=}$${\displaystyle \frac{\sqrt{n}\cancel{\left(\sqrt{m}-\sqrt{n}\right)}}{\cancel{\left(\sqrt{m}-\sqrt{n}\right)}}=}$${\displaystyle \sqrt{n}}$;

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

3) ${\displaystyle \frac{\sqrt{5d}+\sqrt{5b}}{\sqrt{45d}+\sqrt{45b}}=}$${\displaystyle \frac{\cancel{\left(\sqrt{5d}+\sqrt{5b}\right)}}{3\cancel{\left(\sqrt{5d}+\sqrt{5b}\right)}}=}$${\displaystyle \frac{1}{3}}$;

4) ${\displaystyle \frac{\sqrt{cd}+\sqrt{cb}}{\sqrt{ad}+\sqrt{ab}}=}$${\displaystyle \frac{\sqrt{c}\cancel{\left(\sqrt{d}+\sqrt{b}\right)}}{\sqrt{a}\cancel{\left(\sqrt{d}+\sqrt{b}\right)}}=}$${\displaystyle \frac{\sqrt{c}}{\sqrt{a}}}$;

5) ${\displaystyle \frac{\sqrt{x}}{x-\sqrt{x}}=}$${\displaystyle \frac{\cancel{\sqrt{x}}}{\cancel{\sqrt{x}}(\sqrt{x}-1)}=}$${\displaystyle \frac{1}{\sqrt{x}-1}}$;

6) ${\displaystyle \frac{a+5\sqrt{a}}{\sqrt{a}}=}$${\displaystyle \frac{\cancel{\sqrt{a}}(\sqrt{a}+5)}{\cancel{\sqrt{a}}}=}$${\displaystyle \sqrt{a}+5}$;

7) ${\displaystyle \frac{2\sqrt{m}}{m-\sqrt{m}}=}$${\displaystyle \frac{2\cancel{\sqrt{m}}}{\cancel{\sqrt{m}}(\sqrt{m}-1)}=}$${\displaystyle \frac{2}{\sqrt{m}-1}}$;

8) ${\displaystyle \frac{\sqrt{2}x+\sqrt{2}y}{x+y}=}$${\displaystyle \frac{\sqrt{2}\cancel{\left(x+y\right)}}{\cancel{\left(x+y\right)}}=}$${\displaystyle \sqrt{2}}$;

9) ${\displaystyle \frac{\sqrt{8x}+\sqrt{8y}}{\sqrt{2x}+\sqrt{2y}}=}$${\displaystyle \frac{2\cancel{\left(\sqrt{2x}+\sqrt{2y}\right)}}{\cancel{\left(\sqrt{2x}+\sqrt{2y}\right)}}=}$${\displaystyle 2}$;

10) ${\displaystyle \frac{\sqrt{3a}-\sqrt{3b}}{\sqrt{27a}-\sqrt{27b}}=}$${\displaystyle \frac{\cancel{\left(\sqrt{3a}-\sqrt{3b}\right)}}{3\cancel{\left(\sqrt{3a}-\sqrt{3b}\right)}}=}$${\displaystyle \frac{1}{3}}$;

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

12) ${\displaystyle \frac{y-4}{\sqrt{y}+2}=}$${\displaystyle \frac{(\sqrt{y}-2)\cancel{\left(\sqrt{y}+2\right)}}{\cancel{\left(\sqrt{y}+2\right)}}=}$${\displaystyle \sqrt{y}-2}$;

13) ${\displaystyle \frac{\sqrt{m}-5}{m-25}=}$${\displaystyle \frac{\cancel{\left(\sqrt{m}-5\right)}}{\cancel{\left(\sqrt{m}-5\right)}(\sqrt{m}+5)}=}$${\displaystyle \frac{1}{\sqrt{m}+5}}$;

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

15) ${\displaystyle \frac{a^2-5}{a+\sqrt{5}}=}$${\displaystyle \frac{(a-\sqrt{5})\cancel{\left(a+\sqrt{5}\right)}}{\cancel{\left(a+\sqrt{5}\right)}}=}$${\displaystyle a-\sqrt{5}}$;

16) ${\displaystyle \frac{m^2-10}{\sqrt{10}-m}=}$${\displaystyle \frac{\cancel{\left(m-\sqrt{10}\right)}(m+\sqrt{10})}{-\cancel{\left(m-\sqrt{10}\right)}}=}$${\displaystyle -m-\sqrt{10}}$;

17) ${\displaystyle \frac{x-7}{\sqrt{x}-\sqrt{7}}=}$${\displaystyle \frac{\cancel{\left(\sqrt{x}-\sqrt{7}\right)}(\sqrt{x}+\sqrt{7})}{\cancel{\left(\sqrt{x}-\sqrt{7}\right)}}=}$${\displaystyle \sqrt{x}+\sqrt{7}}$;

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

19) ${\displaystyle \frac{m-11}{\sqrt{m}+\sqrt{11}}=}$${\displaystyle \frac{(\sqrt{m}-\sqrt{11})\cancel{\left(\sqrt{m}+\sqrt{11}\right)}}{\cancel{\left(\sqrt{m}+\sqrt{11}\right)}}=}$${\displaystyle \sqrt{m}-\sqrt{11}}$;

20) ${\displaystyle \frac{m}{\sqrt{n}-\sqrt{m}}+\frac{n}{\sqrt{m}-\sqrt{n}}=}$${\displaystyle \frac{m-n}{\sqrt{n}-\sqrt{m}}=}$${\displaystyle \frac{(\sqrt{m}+\sqrt{n})\cancel{\left(\sqrt{m}-\sqrt{n}\right)}}{-\cancel{\left(\sqrt{m}-\sqrt{n}\right)}}=}$${\displaystyle -\sqrt{m}-\sqrt{n}}$;

21) ${\displaystyle \frac{\sqrt{m}}{\sqrt{mn}+n}-\frac{\sqrt{n}}{\sqrt{mn}+m}=}$${\displaystyle \frac{\sqrt{m}}{\sqrt{n}(\sqrt{m}+\sqrt{n})}-\frac{\sqrt{n}}{\sqrt{m}(\sqrt{n}+\sqrt{m})}=}$${\displaystyle \frac{m-n}{\sqrt{mn}(\sqrt{m}+\sqrt{n})}=}$${\displaystyle \frac{(\sqrt{m}-\sqrt{n})\cancel{\left(\sqrt{m}+\sqrt{n}\right)}}{\sqrt{mn}\cancel{\left(\sqrt{m}+\sqrt{n}\right)}}=}$${\displaystyle \frac{\sqrt{m}-\sqrt{n}}{\sqrt{mn}}}$.