Răspuns :

 

[tex]\displaystyle\bf\\Se~da:\\\\tg\,x=3\\\\Se~cere:\\\\r=\frac{4sin^2x+3cos^2x}{2sin^2x+cos^2x}=?\\\\ Simplificam~"fortat"~fractia~cu~~\underline{cos^2x}\\\\\\r=\frac{\dfrac{4sin^2x}{cos^2x}+\dfrac{3cos^2x}{cos^2x}}{\dfrac{2sin^2x}{cos^2x}+\dfrac{cos^2x}{cos^2x}}\\\\\\r=\frac{4\dfrac{sin^2x}{cos^2x}+3\dfrac{cos^2x}{cos^2x}}{2\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{cos^2x}}[/tex]

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[tex]\displaystyle\bf\\r=\frac{4\cdot tg^2x+3\cdot1}{2\cdot tg^2x+1}\\\\\\r=\frac{4\cdot 3^2+3}{2\cdot 3^2+1}\\\\\\r=\frac{4\cdot 9+3}{2\cdot 9+1}\\\\\\r=\frac{36+3}{18+1}\\\\\\\boxed{\bf~r=\frac{39}{19}}[/tex]