Question

Va rogg va rog frumos ajutați maaaaaaa ca habar nu ammm

Answer 1

$2d) \\ {x}^{2012} ( {x}^{2} - x + 1 )$

$2c) \\ = {x}^{m - 1} ( {x}^{2} + x + 1)$

$2b) \\ = xy(y + x + y) = xy(2y + x)$

$2a) \\ = 7( {x}^{2} + {y}^{2} + {z}^{2} )$

$1d) \\ = 21 {x}^{3} \div ( - 7 {x}^{2} ) - 14 {x}^{2} \div (- 7 {x}^{2}) \\ + {x}^{2} + 2x - x - 2 = \\ - 3x + 2 + {x}^{2} + x - 2 = \\ {x}^{2} - 2x$

$1c) \\ \sqrt{3} (2 \sqrt{3} - x ) + \sqrt{6} ( 3\sqrt{2} x + \sqrt{3} ) = \\ 2 \sqrt{9} - \sqrt{3} x + 3 \sqrt{12} x + \sqrt{18} =$

Dar,

$\sqrt{9} = 3 \\ \sqrt{12} = 2 \sqrt{3} \\ \sqrt{18} = 3 \sqrt{2}$

Deci,

$2 \times 3 - \sqrt{3} x + 3 \times 2 \sqrt{3} x + 3 \sqrt{2} = \\ = 6 - \sqrt{3} x + 6 \sqrt{3} x + 3 \sqrt{2} = \\ = 5 \sqrt{3} x + 3 \sqrt{2} + 6$

$1b) \\ (- 72 {x}^{5} + 48 {x}^{4} - 36 {x}^{3} ) \div ( - 4 {x}^{2} ) = \\ - 72 {x}^{5} \div ( - 4 {x}^{2} ) + 48 {x}^{4} \div ( - 4 {x}^{2} )\\ - 36 {x}^{3}\div ( - 4 {x}^{2} ) = \\ 18 {x}^{3} - 12 {x}^{2} + 9x$

$1a) \: {(x + 3)}^{2} - x(x + 1) + 2 = \\ = {x}^{2} + 2 \times x \times 3 + {3}^{2} - {x}^{2} \\ - x + 2 = \\ \\ = {x}^{2} + 6x + 9 - {x}^{2} - x + 2 =$

x^2 se reduce cu - x^2

6x-x=5x

9+2=11

asadar, avem

$= 5x + 11$