Вопрос:

10.12. Найдите значения следующих числовых выражений:

Ответ:

Решение:

а)

\( \frac{(2.708 \cdot 3 + 0.625)}{2.5} = \frac{(8.124 + 0.625)}{2.5} = \frac{8.749}{2.5} = 3.4996 \)
\( \left(1.3 + 0.7(6) + 0.(36)\right) \cdot \frac{110}{173} = \left(1.3 + 4.2 + 0.36\right) \cdot \frac{110}{173} = 5.86 \cdot \frac{110}{173} = \frac{644.6}{173} \approx 3.726 \)

б)

\( \sqrt{7-4\sqrt{3}} \cdot (2+\sqrt{3}) \)
\( \sqrt{7-2\sqrt{12}} = \sqrt{(\sqrt{4}-\sqrt{3})^2} = \sqrt{4}-\sqrt{3} = 2-\sqrt{3} \)
\( (2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4-3=1 \)

в)

\( (\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}})^2 \)
\( \sqrt{11+2\sqrt{18}} = \sqrt{(\sqrt{9}+\sqrt{2})^2} = 3+\sqrt{2} \)
\( \sqrt{11-2\sqrt{18}} = \sqrt{(\sqrt{9}-\sqrt{2})^2} = 3-\sqrt{2} \)
\( (3+\sqrt{2} - (3-\sqrt{2}))^2 = (3+\sqrt{2}-3+\sqrt{2})^2 = (2\sqrt{2})^2 = 4 \cdot 2 = 8 \)

г)

\( \sqrt{2+\sqrt{9+4\sqrt{2}}} \cdot (\sqrt{2}-1) \)
\( \sqrt{9+4\sqrt{2}} = \sqrt{9+2\sqrt{8}} \)
\( \sqrt{9+2\sqrt{8}} = \sqrt{(\sqrt{8}+\sqrt{1})^2} = \sqrt{8}+1 = 2\sqrt{2}+1 \)
\( \sqrt{2+2\sqrt{2}+1} = \sqrt{(\sqrt{2}+1)^2} = \sqrt{2}+1 \)
\( (\sqrt{2}+1)(\sqrt{2}-1) = (\sqrt{2})^2 - 1^2 = 2-1=1 \)

д)

\( (5-3\sqrt{2}) \cdot \sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}} \)
\( \sqrt{9+4\sqrt{2}} = \sqrt{9+2\sqrt{8}} = \sqrt{(\sqrt{8}+1)^2} = \sqrt{8}+1 = 2\sqrt{2}+1 \)
\( 13+30\sqrt{2}+2\sqrt{2}+1 = 14+32\sqrt{2} \)
\( (5-3\sqrt{2}) \sqrt{14+32\sqrt{2}} \)
\( \sqrt{14+32\sqrt{2}} = \sqrt{14+2\sqrt{512}} \)
\( \sqrt{14+2\sqrt{512}} \) - здесь дальнейшие упрощения затруднительны

е)

\( \left(\frac{45!}{43!}-3! \cdot 30\right) \cdot \frac{37!}{2!} \)
\( \frac{45!}{43!} = \frac{45 × 44 × 43!}{43!} = 45 × 44 = 1980 \)
\( 3! × 30 = 6 × 30 = 180 \)
\( 1980 - 180 = 1800 \)
\( \frac{37!}{2!} = \frac{37 × 36 × 35!}{2 × 35!} = \frac{37 × 36}{2} = 37 × 18 = 666 \)
\( 1800 × 666 = 1198800 \)

Ответ: а) 3.4996; (1,3+0,7(6)+0,(36)) × 110/173 ≈ 3.726; б) 1; в) 8; г) 1; д) (5-3√2)√(14+32√2); е) 1198800.

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