101. Выполните умножение: 1) (√99-√44)√11; 2) (4√6-√54+√24). √6; 3) (12-√7)(3+2√7); 4) (2√3+3√5)(3√3 - 2√5); 5) (√14-√10)(√14 + √10); 6) (3√a +7√5) (3√a-7√চ); 7) (√7+1)²; 8) (4/5-5/2)². 102. Упростите выражение: 1) (3√6+5√8-4√32)√2-√108; 2) (√5+7√2)(7√2-√5) – (√10 – 2√5)²; 2 3) (7-√3)² + (4+√3)²; 2 4) (√7-4√3+√7+4√3)². 103. Сократите дробь: 12. x²-11 1) x + √11; 3) a+3√a a-9 m-12√m+36. 5) m-36 2) √x-12. x-144' 17-17 √21-3 4) -; 6) √17 7-√21 104. Освободитесь от иррациональности в знаменателе дроби: 6 17 x-3 5) √x-3 x-4 9) ; √x+5-3 2) 12 √3 1 6) √26-1 x² + 4x 10) ; x+8-2 3) 30 7√5 35 7) √37+√2 x²-16 11) ; 3-√x+5 a³ 4); ba 16 √47-√15 12) √3-x+√3+2x

101. Выполните умножение:

1) (√99-√44)√11;

\[(\sqrt{99} - \sqrt{44}) \cdot \sqrt{11} = (\sqrt{9 \cdot 11} - \sqrt{4 \cdot 11}) \cdot \sqrt{11} = (3\sqrt{11} - 2\sqrt{11}) \cdot \sqrt{11} = \sqrt{11} \cdot \sqrt{11} = 11\]

2) (4√6-√54+√24) ⋅ √6;

\[(4\sqrt{6} - \sqrt{54} + \sqrt{24}) \cdot \sqrt{6} = (4\sqrt{6} - \sqrt{9 \cdot 6} + \sqrt{4 \cdot 6}) \cdot \sqrt{6} = (4\sqrt{6} - 3\sqrt{6} + 2\sqrt{6}) \cdot \sqrt{6} = 3\sqrt{6} \cdot \sqrt{6} = 3 \cdot 6 = 18\]

3) (12-√7)(3+2√7);

\[(12 - \sqrt{7})(3 + 2\sqrt{7}) = 36 + 24\sqrt{7} - 3\sqrt{7} - 2 \cdot 7 = 36 + 21\sqrt{7} - 14 = 22 + 21\sqrt{7}\]

4) (2√3+3√5)(3√3 - 2√5);

\[(2\sqrt{3} + 3\sqrt{5})(3\sqrt{3} - 2\sqrt{5}) = 2\sqrt{3} \cdot 3\sqrt{3} - 2\sqrt{3} \cdot 2\sqrt{5} + 3\sqrt{5} \cdot 3\sqrt{3} - 3\sqrt{5} \cdot 2\sqrt{5} = 6 \cdot 3 - 4\sqrt{15} + 9\sqrt{15} - 6 \cdot 5 = 18 + 5\sqrt{15} - 30 = -12 + 5\sqrt{15}\]

5) (√14-√10)(√14 + √10);

\[(\sqrt{14} - \sqrt{10})(\sqrt{14} + \sqrt{10}) = (\sqrt{14})^2 - (\sqrt{10})^2 = 14 - 10 = 4\]

6) (3√a +7√5) (3√a-7√b);

\[(3\sqrt{a} + 7\sqrt{b})(3\sqrt{a} - 7\sqrt{b}) = (3\sqrt{a})^2 - (7\sqrt{b})^2 = 9a - 49b\]

7) (√7+1)²;

\[(\sqrt{7} + 1)^2 = (\sqrt{7})^2 + 2 \cdot \sqrt{7} \cdot 1 + 1^2 = 7 + 2\sqrt{7} + 1 = 8 + 2\sqrt{7}\]

8) (4√5-5√2)².

\[(4\sqrt{5} - 5\sqrt{2})^2 = (4\sqrt{5})^2 - 2 \cdot 4\sqrt{5} \cdot 5\sqrt{2} + (5\sqrt{2})^2 = 16 \cdot 5 - 40\sqrt{10} + 25 \cdot 2 = 80 - 40\sqrt{10} + 50 = 130 - 40\sqrt{10}\]

102. Упростите выражение:

1) (3√6+5√8-4√32)√2-√108;

\[(3\sqrt{6} + 5\sqrt{8} - 4\sqrt{32}) \cdot \sqrt{2} - \sqrt{108} = (3\sqrt{6} + 5\sqrt{4 \cdot 2} - 4\sqrt{16 \cdot 2}) \cdot \sqrt{2} - \sqrt{36 \cdot 3} = (3\sqrt{6} + 5 \cdot 2\sqrt{2} - 4 \cdot 4\sqrt{2}) \cdot \sqrt{2} - 6\sqrt{3} = (3\sqrt{6} + 10\sqrt{2} - 16\sqrt{2}) \cdot \sqrt{2} - 6\sqrt{3} = (3\sqrt{6} - 6\sqrt{2}) \cdot \sqrt{2} - 6\sqrt{3} = 3\sqrt{12} - 6 \cdot 2 - 6\sqrt{3} = 3\sqrt{4 \cdot 3} - 12 - 6\sqrt{3} = 3 \cdot 2\sqrt{3} - 12 - 6\sqrt{3} = 6\sqrt{3} - 12 - 6\sqrt{3} = -12\]

2) (√5+7√2)(7√2-√5) – (√10 – 2√5)²;

\[(\sqrt{5} + 7\sqrt{2})(7\sqrt{2} - \sqrt{5}) - (\sqrt{10} - 2\sqrt{5})^2 = (7\sqrt{2} + \sqrt{5})(7\sqrt{2} - \sqrt{5}) - ((\sqrt{10})^2 - 2 \cdot \sqrt{10} \cdot 2\sqrt{5} + (2\sqrt{5})^2) = (7\sqrt{2})^2 - (\sqrt{5})^2 - (10 - 4\sqrt{50} + 4 \cdot 5) = 49 \cdot 2 - 5 - (10 - 4\sqrt{25 \cdot 2} + 20) = 98 - 5 - (30 - 4 \cdot 5\sqrt{2}) = 93 - (30 - 20\sqrt{2}) = 93 - 30 + 20\sqrt{2} = 63 + 20\sqrt{2}\]

3) (7-√3)² + (4+√3)²;

\[(7 - \sqrt{3})^2 + (4 + \sqrt{3})^2 = (7^2 - 2 \cdot 7 \cdot \sqrt{3} + (\sqrt{3})^2) + (4^2 + 2 \cdot 4 \cdot \sqrt{3} + (\sqrt{3})^2) = (49 - 14\sqrt{3} + 3) + (16 + 8\sqrt{3} + 3) = 52 - 14\sqrt{3} + 19 + 8\sqrt{3} = 71 - 6\sqrt{3}\]

4) (√7-4√3+√7+4√3)².

\[(\sqrt{7} - 4\sqrt{3} + \sqrt{7} + 4\sqrt{3})^2 = (2\sqrt{7})^2 = 4 \cdot 7 = 28\]

103. Сократите дробь:

1) x²-11 / x + √11;

\[\frac{x^2 - 11}{x + \sqrt{11}} = \frac{(x - \sqrt{11})(x + \sqrt{11})}{x + \sqrt{11}} = x - \sqrt{11}\]

2) √x-12 / x-144;

\[\frac{\sqrt{x} - 12}{x - 144} = \frac{\sqrt{x} - 12}{(\sqrt{x} - 12)(\sqrt{x} + 12)} = \frac{1}{\sqrt{x} + 12}\]

3) a+3√a / a-9;

\[\frac{a + 3\sqrt{a}}{a - 9} = \frac{\sqrt{a}(\sqrt{a} + 3)}{(\sqrt{a} - 3)(\sqrt{a} + 3)} = \frac{\sqrt{a}}{\sqrt{a} - 3}\]

4) 17-17 / √17;

\[\frac{17 - \sqrt{17}}{\sqrt{17}} = \frac{\sqrt{17}(\sqrt{17} - 1)}{\sqrt{17}} = \sqrt{17} - 1\]

5) m-12√m+36 / m-36

\[\frac{m - 12\sqrt{m} + 36}{m - 36} = \frac{(\sqrt{m} - 6)^2}{(\sqrt{m} - 6)(\sqrt{m} + 6)} = \frac{\sqrt{m} - 6}{\sqrt{m} + 6}\]

6) √21-3 / 7-√21

\[\frac{\sqrt{21} - 3}{7 - \sqrt{21}} = - \frac{\sqrt{21} - 3}{\sqrt{21} - 7}\]

104. Освободитесь от иррациональности в знаменателе дроби:

1) 6 / √17;

\[\frac{6}{\sqrt{17}} = \frac{6\sqrt{17}}{17}\]

2) 12 / √3;

\[\frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\]

3) 30 / 7√5;

\[\frac{30}{7\sqrt{5}} = \frac{30\sqrt{5}}{7 \cdot 5} = \frac{6\sqrt{5}}{7}\]

4) a³ / b√a

\[\frac{a^3}{b\sqrt{a}} = \frac{a^3 \sqrt{a}}{b \cdot a} = \frac{a^2 \sqrt{a}}{b}\]

5) x-3 / √x-3;

\[\frac{x - 3}{\sqrt{x} - 3} = \frac{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})}{\sqrt{x} - \sqrt{3}} = \sqrt{x} + \sqrt{3}\]

6) 1 / √26-1;

\[\frac{1}{\sqrt{26} - 1} = \frac{\sqrt{26} + 1}{(\sqrt{26} - 1)(\sqrt{26} + 1)} = \frac{\sqrt{26} + 1}{26 - 1} = \frac{\sqrt{26} + 1}{25}\]

7) 35 / √37+√2

\[\frac{35}{\sqrt{37} + \sqrt{2}} = \frac{35(\sqrt{37} - \sqrt{2})}{(\sqrt{37} + \sqrt{2})(\sqrt{37} - \sqrt{2})} = \frac{35(\sqrt{37} - \sqrt{2})}{37 - 2} = \frac{35(\sqrt{37} - \sqrt{2})}{35} = \sqrt{37} - \sqrt{2}\]

8) 16 / √47-√15;

\[\frac{16}{\sqrt{47} - \sqrt{15}} = \frac{16(\sqrt{47} + \sqrt{15})}{(\sqrt{47} - \sqrt{15})(\sqrt{47} + \sqrt{15})} = \frac{16(\sqrt{47} + \sqrt{15})}{47 - 15} = \frac{16(\sqrt{47} + \sqrt{15})}{32} = \frac{\sqrt{47} + \sqrt{15}}{2}\]

9) x-4 / √x+5-3;

\[\frac{x - 4}{\sqrt{x + 5} - 3} = \frac{(x - 4)(\sqrt{x + 5} + 3)}{(\sqrt{x + 5} - 3)(\sqrt{x + 5} + 3)} = \frac{(x - 4)(\sqrt{x + 5} + 3)}{x + 5 - 9} = \frac{(x - 4)(\sqrt{x + 5} + 3)}{x - 4} = \sqrt{x + 5} + 3\]

10) x² + 4x / √x+8-2;

\[\frac{x^2 + 4x}{\sqrt{x + 8} - 2} = \frac{x(x + 4)(\sqrt{x + 8} + 2)}{(\sqrt{x + 8} - 2)(\sqrt{x + 8} + 2)} = \frac{x(x + 4)(\sqrt{x + 8} + 2)}{x + 8 - 4} = \frac{x(x + 4)(\sqrt{x + 8} + 2)}{x + 4} = x(\sqrt{x + 8} + 2)\]

11) x²-16 / 3-√x+5;

\[\frac{x^2 - 16}{3 - \sqrt{x + 5}} = \frac{(x - 4)(x + 4)(3 + \sqrt{x + 5})}{(3 - \sqrt{x + 5})(3 + \sqrt{x + 5})} = \frac{(x - 4)(x + 4)(3 + \sqrt{x + 5})}{9 - (x + 5)} = \frac{(x - 4)(x + 4)(3 + \sqrt{x + 5})}{4 - x} = - (x + 4)(3 + \sqrt{x + 5})\]

12) x / √3-x+√3+2x.

\[\frac{x}{\sqrt{3 - x} + \sqrt{3 + 2x}} = \frac{x(\sqrt{3 - x} - \sqrt{3 + 2x})}{(\sqrt{3 - x} + \sqrt{3 + 2x})(\sqrt{3 - x} - \sqrt{3 + 2x})} = \frac{x(\sqrt{3 - x} - \sqrt{3 + 2x})}{3 - x - (3 + 2x)} = \frac{x(\sqrt{3 - x} - \sqrt{3 + 2x})}{-3x} = - \frac{\sqrt{3 - x} - \sqrt{3 + 2x}}{3} = \frac{\sqrt{3 + 2x} - \sqrt{3 - x}}{3}\]