Задача 45142 ...
Условие
1)√1200–20√2,43+4,5√0,48
2)√20/81–2/9√45/49–11/20√80/121
Решение
1200=400·3
√1200=√400·3=√400·√3=20√3
√2,43=[m]\sqrt{\frac{243}{100}}=\sqrt{\frac{81\cdot 3}{100}}[/m]=[m]\frac{9}{10}[/m]√3
√0,48=[m]\sqrt{\frac{48}{100}}=\sqrt{\frac{16\cdot 3}{100}}[/m]=[m]\frac{4}{10}[/m]√3
√1200–20√2,43+4,5√0,48=20√3–18√3+1,8√3=3,8√3
2)
[m]\sqrt{\frac{20}{81}}-\frac{2}{9}\sqrt{\frac{45}{49}}-\frac{11}{20}\sqrt{\frac{80}{121}}=\sqrt{\frac{4\cdot 5}{81}}-\frac{2}{9}\sqrt{\frac{5\cdot 9}{49}}-\frac{11}{20}\sqrt{\frac{16\cdot 5}{121}}=[/m]
[m]=\frac{2}{9}\sqrt{5}-\frac{2}{9}\cdot \frac{3}{7}\sqrt{5}-\frac{11}{20}\cdot \frac{4}{11}\sqrt{5}=\frac{2}{9}\cdot \frac{4}{7}\sqrt{5}-\frac{1}{5}\sqrt{5}=-\frac{23}{315}\sqrt{5}[/m]