Question

fie numerele reale a=(4/√3-7/√2)*√6+|12-7√3| si b=√6²+8²-8/√2. a) Aratati ca a=4(√2-3). b) Calculati (a+b)¹⁰. Va rog am nevoie urgent!!​

Answer 1

Explicație pas cu pas:

a = (4/√3 - 7/√2)*√6 + |12 - 7√3|

b = √(6² + 8²) - 8/√2

a) 12 < 7√3 => |12 - 7√3| = 7√3 - 12

$a = \left( \frac{4}{ \sqrt{3}} - \frac{7}{ \sqrt{2}}\right) \times \sqrt{6} + |12 - 7 \sqrt{3} | \\ = \frac{4 \sqrt{6} }{ \sqrt{3} } - \frac{7 \sqrt{6} }{\sqrt{2} } + 7 \sqrt{3} - 12 \\ = 4 \sqrt{2} - 7 \sqrt{3} + 7 \sqrt{3} - 12 \\ = 4 \sqrt{2} - 12 = 4( \sqrt{2} - 3)$

b)

$b = \sqrt{ {6}^{2} + {8}^{2} } - \frac{8}{ \sqrt{2} } = \sqrt{36 + 64} - \frac{8 \sqrt{2} }{2} \\ = \sqrt{100} - 4 \sqrt{2} = 10 - 4 \sqrt{2}$

$a + b = 4 \sqrt{2} - 12 + 10 - 4 \sqrt{2} = - 2$

${(a + b)}^{10} = {( - 2)}^{10} = {2}^{10} = 1024$