Question

ma puteti ajuta ,va rog frumos

Answer 1

a)[tex] \sqrt{2x-1}=3;2x-1=9;2x=10;x=5 \\ b) \sqrt[3]{2x-1}=3;2x-1=27;2x=28;x=14 \\ c) \sqrt{2x+3}=4x;x \ \textgreater \ 0;2x+3=16 x^{2} ;16 x^{2} -2x-3=0 ;x= \frac{1}{2 \\ } \\ d) \sqrt[3]{x+2 \sqrt{2} }=2 \sqrt{2} ;x+2 \sqrt{2}=2 \sqrt{2};x=0 \\ e) \sqrt{3-x}=-2x;x\ \textless \ 0;3-x=4 x^{2} ;4 x^{2} +x-3=0;x=-1 \\ f) \sqrt{ x^{2} -3}=3-x;x\ \textless \ 3; x^{2} -3=9-6x+ x^{2} ;6x=12;x=2 \\ g) \sqrt[3]{x^3-7}=x-1;x^3-7=x^3-3 x^{2} +3x-1 ;3 x^{2} -3x-6=0 [/tex]$x=-1;x=2 \\ h) \sqrt{ x^{2}-3x+2 }=1-x;x \leq 1; x^{2} -3x+2= 1-2x+x^{2} ;x=1$