Question

Sa se rezolve ecuatiile

Answer 1

$\displaystyle\bf\\1)~\sqrt{x-1}=6~\bigg |^2 \implies |x-1|=36\Leftrightarrow x-1=36,~evident~\boxed{\bf x =37}~.\\2)~\sqrt[3]{3x+1} =3~\bigg |^3 \implies 3x+1=27 \Leftrightarrow 3x=26 \implies \boxed{\bf x=\frac{26}{3}}~.\\\\3)~\sqrt{x+12} =x+5~\bigg |^2 \implies x+12=(x+5)^2 \Leftrightarrow x+12=x^2 + 10x+25 \Leftrightarrow \\x=x^2+10x+13 \Leftrightarrow x^2+9x+13=0 \implies \boxed{\\\bf x=\frac{-9\pm\sqrt{81-52} }{2} =\frac{-9+\sqrt{29}}{2}}~.$$\displaystyle\bf\\4)~\sqrt[3]{x^3+8} =x+2~\bigg |^3 \Leftrightarrow x^3+8=(x+2)^3 \Leftrightarrow (x+2)(x^2-2x+4)-(x+2)^3 \Leftrightarrow\\(x+2)(x^2-2x+4-(x+2)^2)=0.\\cazul~1~:~x+2=0 \implies \boxed{\bf x=-2}~.\\cazul~2~:~x^2-2x+4-(x+2)^2=0 \Leftrightarrow x^2-2x+4-x^2-4x-4=0 \Leftrightarrow\\-6x=0 \implies \boxed{\bf x=0}~.\\\\5)~\sqrt{x+4} + \sqrt{2x+6} =7 \Leftrightarrow \sqrt{x+4} = 7-\sqrt{2x+6}~\bigg |^2 \\\implies x+4=49-14\sqrt{x} +2x+6 \Leftrightarrow x+4=55-14\sqrt{2x+6} +2x \Leftrightarrow$

$\displaystyle\bf\\14\sqrt{2x+6}=51+x \bigg |^2 \implies 290x-1425-x^2=0 \Letrightarrow -(x-5)(x-285)=0.\\cazul~1~:x-5=0 \implies x=5.\\cazul~2~:x-285=0 \implies x=285 \implies nu~se~verifica.\\ deci~~\boxed{\bf x=5}~.$

Answer 2

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