Odpowiedź:
DF
Szczegółowe wyjaśnienie:
A)
[tex]2-\sqrt3x=5\sqrt3+2\\-\sqrt3x=5\sqrt3\ |:(-\sqrt3)\\x=-5[/tex]
B)
[tex]3\sqrt3+\frac{x}{\sqrt3}=2\sqrt3\ |*\sqrt3\\3*3+x=2*3\\9+x=6\\x=-3[/tex]
C)
[tex]2x-\sqrt3=1-2\sqrt3x\\2x+2\sqrt3x=1+\sqrt3\\2x(1+\sqrt3)=1+\sqrt3\ |:(1+\sqrt3)\\2x=1\ |:2\\x=\frac{1}{2}[/tex]
D)
[tex]\sqrt3(x+1)=2+\sqrt3\\\sqrt3x+\sqrt3=2+\sqrt3\\\sqrt3x=2\ |:\sqrt3\\x=\frac{2}{\sqrt3}*\frac{\sqrt3}{\sqrt3}\\x=\frac{2\sqrt3}{3}[/tex]
E)
[tex]\sqrt3x+2\sqrt3=2+x\\\sqrt3x-x=2-2\sqrt3\\x(\sqrt3-1)=-2(\sqrt3-1)\ |:(\sqrt3-1)\\x=-2[/tex]
F)
[tex]\sqrt3(x-1)+2x=-2\\\sqrt3x-\sqrt3+2x=-2\\\sqrt3x+2x=\sqrt3-2\\x(\sqrt3+2)=\sqrt3-2\ |:(\sqrt3+2)\\x=\frac{\sqrt3-2}{\sqrt3+2}*\frac{\sqrt3-2}{\sqrt3-2}\\x=\frac{(\sqrt3-2)^2}{(\sqrt3)^2-2^2}\\x=\frac{(\sqrt3)^2-2*\sqrt3*2+2^2}{3-4}\\x=\frac{3-4\sqrt3+4}{-1}\\x=\frac{7-4\sqrt3}{-1}\\x=4\sqrt3-7[/tex]
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Odpowiedź:
DF
Szczegółowe wyjaśnienie:
A)
[tex]2-\sqrt3x=5\sqrt3+2\\-\sqrt3x=5\sqrt3\ |:(-\sqrt3)\\x=-5[/tex]
B)
[tex]3\sqrt3+\frac{x}{\sqrt3}=2\sqrt3\ |*\sqrt3\\3*3+x=2*3\\9+x=6\\x=-3[/tex]
C)
[tex]2x-\sqrt3=1-2\sqrt3x\\2x+2\sqrt3x=1+\sqrt3\\2x(1+\sqrt3)=1+\sqrt3\ |:(1+\sqrt3)\\2x=1\ |:2\\x=\frac{1}{2}[/tex]
D)
[tex]\sqrt3(x+1)=2+\sqrt3\\\sqrt3x+\sqrt3=2+\sqrt3\\\sqrt3x=2\ |:\sqrt3\\x=\frac{2}{\sqrt3}*\frac{\sqrt3}{\sqrt3}\\x=\frac{2\sqrt3}{3}[/tex]
E)
[tex]\sqrt3x+2\sqrt3=2+x\\\sqrt3x-x=2-2\sqrt3\\x(\sqrt3-1)=-2(\sqrt3-1)\ |:(\sqrt3-1)\\x=-2[/tex]
F)
[tex]\sqrt3(x-1)+2x=-2\\\sqrt3x-\sqrt3+2x=-2\\\sqrt3x+2x=\sqrt3-2\\x(\sqrt3+2)=\sqrt3-2\ |:(\sqrt3+2)\\x=\frac{\sqrt3-2}{\sqrt3+2}*\frac{\sqrt3-2}{\sqrt3-2}\\x=\frac{(\sqrt3-2)^2}{(\sqrt3)^2-2^2}\\x=\frac{(\sqrt3)^2-2*\sqrt3*2+2^2}{3-4}\\x=\frac{3-4\sqrt3+4}{-1}\\x=\frac{7-4\sqrt3}{-1}\\x=4\sqrt3-7[/tex]