a) x∈(1,5)
[tex](x+1)^2+4 < (2+x)^2-(2-x)^2\\x^2+2x+1+4 < 4+4x+x^2-(4-4x+x^2)\\x^2+2x+5 < 4+4x+x^2-4+4x-x^2\\x^2+2x+5 < 8x\\x^2-6x+5 < 0\\\\\Delta=b^2-4ac\\\Delta=36-4*5*1=36-20=16\\\sqrt{\Delta} =4\\\\x_{1} =\frac{-b-\sqrt{\Delta}}{2a} =\frac{6-4}{2}=1\\ \\x_{2} =\frac{-b-\sqrt{\Delta}}{2a} =\frac{6+4}{2}=5[/tex]
b) x = 5
[tex]x^3-5x^2=10-2x\\x^3-5x^2+2x-10=0\\x^2(x-5)+2(x-5)=0\\x=5[/tex]
c) x = 2 , x = 1 i 1/2
[tex]\frac{x+1}{4-x} =\frac{x+4}{10-3x} \\\\zalozenia:\\x\neq 4 \land x\neq\frac{10}{3} \\x\in R \backslash[3\frac{1}{3},4 ][/tex]
[tex]\frac{x+1}{4-x} =\frac{x+4}{10-3x}/*(4-x),*(10-3x)\\\\(x+1)*(10-3x)=(4+x)*(4-x)\\\\10x-3x^2+10-3x=16-x^2\\\\-3x^2+7x+10=16-x^2/+x^2,-16\\\\-2x^2+7x-6=0\\\\\Delta=49-4*(-2)*(-6)=1\\\sqrt{\Delta}=1 \\\\x_{1} =\frac{-7-1}{-4} =2\\\\x_{2} =\frac{-7+1}{-4} =1\frac{1}{2}[/tex]
