2. Rozwiąż równanie.b) [tex](3 -x )^{2} - (x+ \frac{1}{3})^{2} = \frac{2}{9} [/tex]d) [tex]4(x + 2)^.... Question from @lkukawka

June 2022 1 3 Report

2. Rozwiąż równanie.
b)
[tex](3 -x )^{2} - (x+ \frac{1}{3})^{2} = \frac{2}{9} [/tex]
d)
[tex]4(x + 2)^{2}-(2x - 1)^{2}=20x+10[/tex]

3. Rozwiąż nierówność.
[tex]c) \: 2(x + 2)^{2} -( \sqrt{2}x - 2)^{2} \geqslant 0[/tex]
Daje naj

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Verified answer

Zadanie 2

$(3 - x) {}^{2} - (x + \frac{1}{3} ) {}^{2} = \frac{2}{9} \\ \\ 9 - 6x + {x}^{2} - ({x}^{2} - \frac{2}{3} x - \frac{1}{9}) = \frac{2}{9} \\ \\ 9 - 6x + {x}^{2} - {x}^{2} - \frac{2}{3} x - \frac{1}{9} = \frac{2}{9} \\ \\ \frac{80}{9} - \frac{20}{3} x = \frac{2}{9} \\ \\ 80 - 60x = 2 \\ - 60x = 2 - 80 \\ - 60x = - 78/ \div ( - 60) \\ x = \frac{13}{10} = 1\frac{3}{10} = 1.3$

$4(x + 2) {}^{2} - (2x - 1) {}^{2} = 20x + 10 \\ 4( {x}^{2} + 4x + 4) - (4 {x}^{2} - 4x + 1) = 20x + 10 \\ 4 {x}^{2} + 16x + 16 - 4 {x}^{2} + 4x - 1 = 20x + 10 \\ 20x + 15 = 20x + 10 \\ 15 = 10$

Równanie nie ma rozwiązania.

Zadanie 3

$2(x + 2) {}^{2} - ( \sqrt{2} x - 2) {}^{2} \geqslant 0 \\ 2 ( {x}^{2} + 4x + 4) - (2 {x}^{2} - 4 \sqrt{2} x + 4) \geqslant 0 \\ 2 {x}^{2} + 8x + 8 - 2 {x}^{2} + 4 \sqrt{2} x - 4 \geqslant 0 \\ 8x + 4 + 4 \sqrt{2}x \geqslant 0 \\ 8x + 4 \sqrt{2} x \geqslant - 4 \\ (8 + 4 \sqrt{2} )x \geqslant - 4 \\ \\ x \geqslant \frac{4}{ - 8 + 4 \sqrt{2} } \\ x \geqslant - \frac{4}{4(2 + \sqrt{2} )} \\ x \geqslant - \frac{1}{2 + \sqrt{2} } \\ x \geqslant - \frac{2 - \sqrt{2} }{2}$