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Proszę o pomoc w tym zadaniu będę mega wdzięczna zadanie w załączniku
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Proszę O Pomoc W Tym Zadaniu Będę Mega Wdzięczna Zadanie W Załączniku PS Daje Naj I class=

Odpowiedź :

Basetlaviz

[tex]a)\\x^{2}-4x+4 = 0\\\\(x-2)^{2} = 0\\\\x-2 = 0\\\\x_{o} = 2[/tex]

[tex]b)\\4x^{2}+4x+1 = 0\\\\(2x+1)^{2} = 0\\\\2x+1 = 0\\\\2x = -1\\\\x_{o} = -\frac{1}{2}[/tex]

[tex]c)\\x^{2}+10x+25 = 0\\\\(x+5)^{2} = 0\\\\x+5 = 0\\\\x_{o} = -5[/tex]

[tex]d)\\x^{2}-2\sqrt{3}x + 3 = 0\\\\(x-\sqrt{3})^{2} = 0\\\\x-\sqrt{3} = 0\\\\x_{o} = \sqrt{3}[/tex]

[tex]e)\\4x^{2}+4\sqrt{5}+5 = 0\\\\(2x+\sqrt{5})^{2} = 0\\\\2x+\sqrt{5} = 0\\\\2x = -\sqrt{5} \ \ /:2\\\\x_{o} = -\frac{\sqrt{5}}{2}}[/tex]

[tex]f)\\5x^{2}+2\sqrt{5}x+1 = 0\\\\(\sqrt{5}x + 1)^{2} = 0\\\\\sqrt{5}x+1 = 0\\\\\sqrt{5}x = -1 \ \ /:\sqrt{5}\\\\x = -\frac{1}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}\\\\x_{o}=-\frac{\sqrt{5}}{5}[/tex]

Shookiviz

Odpowiedź:

[tex]a) x^{2} -4x+4=0\\(x-2)^{2} =0\\x-2=0\\x=2\\b)4x^{2} +4x+1=0\\(2x+1)^{2} =0\\2x+1=0\\2x=-1\\x=-\frac{1}{2} \\c) x^{2} +10x+25=0\\(x+5)^{2} =0\\x+5=0\\x=-5\\d) x^{2} -2\sqrt{3} x+3=0\\(x-\sqrt{3} )^{2}=0\\(x-\sqrt{3} )=0\\x=\sqrt{3} \\e)4x^{2} +4\sqrt{5} +5=0\\(2x+\sqrt{5} )^{2} =0\\2x+\sqrt{5} =0\\2x=-\sqrt{5} \\x=-\frac{\sqrt{5} }{2} \\f) 5x^{2} +2\sqrt{5} x+1=0\\(\sqrt{5} x+1)^{2} =0\\\sqrt{5} x+1=0\\\sqrt{5} x=-1\x=\frac{-1}{\sqrt{5} } *\frac{\sqrt{5} }{\sqrt{5}}\\[/tex]

[tex]x=- \frac{\sqrt{5} }{5}[/tex]

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