Question

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Answer 1

a)

a= 201/2 + 401/4 + 601/6 + 1201/12

Numitor comun: 12

=> a = 1206/12 + 1203/12 + 1202/12 + 1201/12

= (1206+1203+1202+1201)/12 = 4812/12 = 401

b)

[(√2+√3+√32+√48)/(√2-√3+√32-√48)] ÷ (√2+√3)/(√2-√3)

(√2+√3+√32+√48) = √2+√3+4√2+4√3 = 5√2+5√3

(√2-√3+√32-√48) = √2-√3+4√2-4√3 = 5√2-5√3

<=>

$\frac{5 \sqrt{2} + 5 \sqrt{3} }{5 \sqrt{2 } - 5 \sqrt{3} } \div \frac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

$= \frac{5( \sqrt{2} + \sqrt{3}) }{5( \sqrt{2} - \sqrt{3}) } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$

$= \frac{5( \sqrt{2} + \sqrt{3})( \sqrt{2} - \sqrt{3}) }{5( \sqrt{2} - \sqrt{3})( \sqrt{2} + \sqrt{3}) }$

$= \frac{5(2 - 3)}{5(2 - 3)} = \frac{ - 5}{ - 5} = 1$

Media aritmetică: (a+b)/2 = (401+1)/2 = 402/2 = 201

Answer 2

Explicație pas cu pas:

a)

$\frac{1206 + 1203 + 1202 + 1201}{12} = \frac{4812}{12} = 401$

b)

$\frac{ \sqrt{2} + \sqrt{3} + 4 \sqrt{2} + 4 \sqrt{3} }{ \sqrt{2} - \sqrt{3} + 4 \sqrt{2} - 4 \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} } = \frac{5 \sqrt{2} + 5 \sqrt{3} }{5 \sqrt{2} - 5 \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} } = \frac{5( \sqrt{2} + \sqrt{3}) }{5( \sqrt{2} - \sqrt{3} ) } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} } = \frac{1}{1} = 1$

$m = \frac{a + b}{2} = \frac{401 + 1}{2} = \frac{402}{2} = 201$