Odpowiedź:
Szczegółowe wyjaśnienie:
a.
[tex]\frac{x}{4} -\frac{5-x}{6} =\frac{2}{3}[/tex] /*12
D=R
3*x-2(5-x)=4*2
3x-10+2x=8
5x=8+10
5x=18
[tex]x=\frac{18}{5} =3\frac{3}{5}[/tex]
b.
[tex]2x=\sqrt{5} (x+1)[/tex]
[tex]2x=\sqrt{5} x+\sqrt{5} \\2x-\sqrt{5} x=\sqrt{5} \\(2-\sqrt{5} )x=\sqrt{5} \\x=\frac{\sqrt{5} }{2-\sqrt{5} } \\x=\frac{5*(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} =\frac{10+5\sqrt{5} }{4-5} \\x=-10-5\sqrt{5}[/tex]
c.
[tex]\frac{x-2}{x+4} -\frac{5}{6} =0[/tex]
D: x≠-4
D=R\{-4}
[tex]\frac{x-2}{x+4}=\frac{5}{6} \\6(x-2)=5(x+4)\\6x-12=5x+20\\6x-5x=20+12\\x=32[/tex]∈D
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Odpowiedź:
Szczegółowe wyjaśnienie:
a.
[tex]\frac{x}{4} -\frac{5-x}{6} =\frac{2}{3}[/tex] /*12
D=R
3*x-2(5-x)=4*2
3x-10+2x=8
5x=8+10
5x=18
[tex]x=\frac{18}{5} =3\frac{3}{5}[/tex]
b.
[tex]2x=\sqrt{5} (x+1)[/tex]
D=R
[tex]2x=\sqrt{5} x+\sqrt{5} \\2x-\sqrt{5} x=\sqrt{5} \\(2-\sqrt{5} )x=\sqrt{5} \\x=\frac{\sqrt{5} }{2-\sqrt{5} } \\x=\frac{5*(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} =\frac{10+5\sqrt{5} }{4-5} \\x=-10-5\sqrt{5}[/tex]
c.
[tex]\frac{x-2}{x+4} -\frac{5}{6} =0[/tex]
D: x≠-4
D=R\{-4}
[tex]\frac{x-2}{x+4}=\frac{5}{6} \\6(x-2)=5(x+4)\\6x-12=5x+20\\6x-5x=20+12\\x=32[/tex]∈D