Question

Proszę o pomoc , rozwiąż równanie, korzystając ze wzorów na różnicę kwadratów ,z góry dziękuję

Answer 1

Verified answer

[tex]a)\\\\16x^2-1=0\\\\(4x-1)(4x+1)=0\\\\4x-1=0\ \ \ \ \vee\ \ \ \ 4x+1=0\\\\4x=1\ \ |:4\ \ \vee\ \ \ \ 4x=-1\ \ |:4\\\\x=\frac{1}{4}\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{1}{4}\\\\\\b)\\\\25x^2-1=0\\\\(5x-1)(5x+1)=0\\\\5x-1=0\ \ \ \ \vee\ \ \ \ 5x+1=0\\\\5x=1\ \ |:5\ \ \vee\ \ \ \ 5x=-1\ \ |:5\\\\x=\frac{1}{5}\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{1}{5}[/tex]

[tex]c)\\\\1-\frac{9}{4}x^2=0\\\\(1-\frac{3}{2}x)(1+\frac{3}{2}x)=0\\\\1-\frac{3}{2}x=0\ \ \ \ \vee\ \ \ \ 1+\frac{3}{2}x=0\\\\-\frac{3}{2}x=-1\ \ |:(-\frac{3}{2})\ \ \ \ \vee\ \ \ \ \frac{3}{2}x=-1\ \ |:\frac{3}{2}\\\\x=1\cdot\frac{2}{3}\ \ \ \ \vee\ \ \ \ x=-1\cdot\frac{2}{3}\\\\x=\frac{2}{3}\ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{2}{3}[/tex]

[tex]d)\\\\25-4x^2=0\\\\(5-2x)(5+2x)=0\\\\5-2x=0\ \ \ \ \vee\ \ \ \ 5+2x=0\\\\-2x=-5\ \ |:(-2)\ \ \ \ \vee\ \ \ \ 2x=-5\ \ |:2\\\\x=\frac{5}{2}\ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{5}{2}\\\\\\e)\\\\3-2x^2=0\\\\(\sqrt{3}-\sqrt{2}x)(\sqrt{3}+\sqrt{2}x)=0\\\\\sqrt{3}-\sqrt{2}x=0\ \ \ \ \vee\ \ \ \ \sqrt{3}+\sqrt{2}x=0\\\\-\sqrt{2}x=-\sqrt{3}\ \ |:(-\sqrt{2})\ \ \ \ \vee\ \ \ \ \sqrt{2}x=-\sqrt{3}\ \ |:\sqrt{2}\\\\x=\frac{\sqrt{3}}{\sqrt{2}}\ \ \ \ \ \ \vee\ \ \ \ x=-\frac{\sqrt{3}}{\sqrt{2}}[/tex]

[tex]x=\frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\ \ \ \ \vee\ \ \ \ x=-\frac{\sqrt{3} }{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\\\x=\frac{\sqrt{6}}{2}\ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{\sqrt{6}}{2}[/tex]

[tex]f)\\\\9-36x^2=0\ \ |:9\\\\1-4x^2=0\\\\(1-2x)(1+2x)=0\\\\1-2x=0\ \ \ \ \vee\ \ \ \ 1+2x=0\\\\-2x=1\ \ |:(-2)\ \ \ \ \vee\ \ \ \ 2x=1\ \ |:2\\\\x=-\frac{1}{2}\ \ \ \ \ \ \vee\ \ \ \ x=\frac{1}{2}[/tex]