[tex] 4x^2+4\sqrt{ 2 }x+3 \leq 4x+2\sqrt{ 2 }\\\\4x^2+4\sqrt{ 2 }x+3 -4x-2\sqrt{ 2 }\leq 0\\\\4x^2+4\sqrt{2}x-4x+3 +2\sqrt{2}\jscaleq 0\\\\4x^2+(4\sqrt{2}-4)+3-2\sqrt{2}\leq 0\\\\miejsca\ zerowe:\\\\4x^2+(4\sqrt{2}-4)+3-2\sqrt{2}=0\\\\a=4,\ \ b=(4\sqrt{2}-4),\ \ c=3-2\sqrt{2}[/tex]
[tex]\Delta =b^2-4ac=(4\sqrt{2}-4)^2-4*4*(3-2\sqrt{2})=32-32\sqrt{2}+16 -48+32\sqrt{2}=0\\\\x_{0}=\frac{-(4\sqrt{2}-4)}{8}=\frac{- 4\sqrt{2}+4 }{8}=\frac{ \not{4}^1 (1-\sqrt{2})}{\not{8}^2}=\frac{1-\sqrt{2}}{2} \\\\a>0 ,\ ramiona\ skierowane\ do\ gory ,\ \ \Delta =0\ jedno\ miejsce\ zerowe , ktore\ lezy\ na\ osi\ OX , \\\\rozwiazaniem\ tej\ nierownosci\ jest\ :\\\\x\in\left\{\frac{ 1-\sqrt{2}}{2}\right\} [/tex]
