Odpowiedź:
7
[tex] | \frac{1}{2}x - 1 | \leqslant 2 \\ 0.5x - 1 \leqslant 2 \\ 0.5x \leqslant 3 \\ x \leqslant 6 \\ i \\ - ( 0.5x - 1) \leqslant - 2 \\ 0.5x \geqslant - 1 \\ x \geqslant - 2 [/tex]
x=[-2,6]
b)
[tex] \sqrt{ {x}^{2} - 6x + 9} > 4 \\ \sqrt{ {(x - 3)}^{2} } > 4 \\ |x - 3| > 4 \\ x > 7 \\ i \\ x < - 1 \\ x = ( - \infty . - 1)u(7. \infty )[/tex]
6
[tex] \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} = \frac{ {( \sqrt{3} - 1) }^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1)} = \frac{3 - 2 \sqrt{3} + 1}{3 - 1} = \frac{4 - 2 \sqrt{3} }{2} = \frac{2(2 - \sqrt{3} )}{2} = 2 - \sqrt{3} [/tex]
5
[tex]4 {(x + 2)}^{2} - (2x - 1)(2x + 1) = 9 \\ 4 {x}^{2} + 16x + 16 - 4 {x}^{2} + 1 = 9 \\ 16x + 17 = 9 \\ 16x = - 8 \\ x = - 0.5[/tex]
4
[tex]a \\ \\ \sqrt{( \sqrt{2} - 1)( \sqrt{2} + 1) } = \sqrt{2 - 1} = \sqrt{1} = 1 \\ \\ b \\ \\ {( \sqrt{2 + \sqrt{5} } + \sqrt{2 - \sqrt{5} } ) }^{2} = {( \sqrt{2 + \sqrt{5} }) }^{2} + 2( \sqrt{2 + \sqrt{5} } )( \sqrt{2 - \sqrt{5} } ) + {( \sqrt{2 - \sqrt{5} })}^{2} = \\ = 2 + \sqrt{5} + 2([/tex]
