Question

Efectuați:

$a) \\ \\ \sqrt{48} \times (2 \sqrt{75} - \sqrt{27} + 2 \sqrt{27} ) - \sqrt{48} \times \sqrt{12} \\ \\ b) \\ \\ \sqrt{243} \times \sqrt{147} - 2 \sqrt{3} \times 7 \sqrt{3} + 3 \sqrt{8}(5 \sqrt{72} + 6 \sqrt{128})$

Trebuie cu rezolvare!!!

Answer 1

a) √48 × ( 2√75 - √27 + 2√27 ) - √48 × √12 =
= 4√3 × ( 10√3 - 3√3 + 6√3 ) - √576 =
= 4√3 × 13√3 - 24 =
= 4 × 3 × 13 - 24 =
= 156 - 24 = 132

b) √243 × √147 - 2√3 × 7√3 + 3√8 × ( 5√72 + 6√128 ) =
= √35721 - 2 × 3 × 7 + 3 × 2√2 × ( 30√2 + 48√2 ) =
= 189 - 42 + 3 × 2√2 × 78√2 =
= 189 - 42 + 3 × 2 × 2 × 78 =
= 189 - 42 + 936 =
= 1125 - 42 = 1083 

Answer 2

a)

[tex]\it \sqrt{48} = \sqrt{16\cdot3} = 4\sqrt3 \\ \\ \sqrt{75} = \sqrt{25\cdot3} = 5\sqrt3 \\ \\ \sqrt27 = \sqrt{9\cdot3}= 3\sqrt3 \\ \\ \sqrt{12} = \sqrt{4\cdot3} = 2\sqrt3[/tex]

Expresia din enunț devine:

[tex]\it 4\sqrt3(10\sqrt3-3\sqrt3+6\sqrt3) - 4\sqrt3\cdot2\sqrt3 = 4\sqrt3\cdot13\sqrt3- 4\sqrt3 \cdot 2\sqrt3=\\ \\ = 4\sqrt3(13\sqrt3 -2\sqrt3) = 4\sqrt3\cdot11\sqrt3 = 44\cdot3 = 132 [/tex]