[tex]a)\\\\\frac{ 3x}{ 4x^2-25} =\frac{ x-1}{2x-5 } \\\\ \frac{ 3x}{(2x-5)(2x+5)} =\frac{ x-1}{2x-5 } \\\\2x-5\neq 0\ \ lub\ \ 2x+5\neq 0\\\\ 2x \neq 5\ \ lub\ \ 2x \neq -5 \\\\ x \neq \frac{5}{2}\ \ lub\ \ x \neq -\frac{5}{2}\\\\D=R\setminus \left\{-\frac{5}{2},\frac{5}{2} \right\} \\[/tex]
[tex]\frac{ 3x}{(2x-5)(2x+5)} =\frac{ x-1}{2x-5 }\\\\3x\not{(2x-5)}^1=\not{(2x-5)}^1(2x+5)(x-1) \\\\3x=(2x+5)(x-1) \\\\3x=2x^2-2x+5x-5\\\\2x^2+3x-3x+5=0\\\\2x^2 +5=0\\\\a=2,\ \ b=0,\ \ c=5\\\\\Delta =b^2-4ac=0^2-4*2* (-5)=40\\\\\sqrt{\Delta }=\sqrt{40}= \sqrt{4*10}=2\sqrt{10}\\\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{0 -2\sqrt{10}}{2*2}=\frac{-2\sqrt{10}}{4}= -\frac{\sqrt{10}}{2}[/tex]
[tex]x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{0+2\sqrt{10}}{2*2}=\frac{ 2\sqrt{10}}{4}= \frac{\sqrt{10}}{2}\\\\x\in\left\{ -\frac{\sqrt{10}}{2},\frac{\sqrt{10}}{2}\right\}[/tex]
[tex]b)\\\\\frac{3}{ x-2} =\frac{2}{x+2 }+\frac{ 7}{ x^2-4}\\\\\frac{3}{ x-2} =\frac{2}{x+2 }+\frac{ 7}{ (x -2)(x+2)}\\\\x-2\neq 0\ \ lub\ \ x+2\neq 0\\\\x\neq 2\ \ lub\ \ x\neq -2\\\\D=R\setminus \left\{ -2,2\right\}[/tex]
[tex]\frac{3}{ x-2} = \frac{2(x-2)+7}{ (x -2)(x+2)}\\\\\frac{3}{ x-2} = \frac{2 x-4 +7}{ (x -2)(x+2)} \\\\\frac{3}{ x-2} = \frac{2 x+3}{ (x -2)(x+2)}[/tex]
[tex]3(x +2)\not{(x-2)}^1 = \not{(x-2)}^1(2x+3)\\\\3(x+2)=2x+3\\\\3x+6=2x+3\\\\3x-2x=3-6\\\\x=-3[/tex]
[tex]c)\\\\\frac{4}{x^2-3x-10}= \frac{x}{x-5}+\frac{x-1}{x+2}\\\\\frac{4}{ (x-5)(x+2)}= \frac{x}{x-5}+\frac{x-1}{x+2}\\\\x-5\neq 0\ \ lub\ \ x+2\neq 0\\\\x \neq 5\ \ lub\ \ x \neq -2\\\\D=R\left\{ -2,5\right\}[/tex]
[tex]\frac{4}{ (x-5)(x+2)}= \frac{x(x+2)+(x-5)(x-1)}{(x-5)(x+2)} \\\\\frac{4}{ (x-5)(x+2)}= \frac{x^2+2x+ x^2-x-5x+5}{(x-5)(x+2)}\\\\ \frac{4}{ (x-5)(x+2)}= \frac{2x^2-4x+5}{(x-5)(x+2)}[/tex]
[tex]2x^2-4x+5=4\\\\2x^2-4x+5-4=0\\\\2x^2-4x+1=0\\\\a=2,\ \ b=-4,\ \ c=1\\\\\Delta =b^2-4ac=(-4)^2-4*2*1=16-8=8\\\\\sqrt{\Delta }=\sqrt{8}=\sqrt{4*2}=2\sqrt{2}[/tex]
[tex]2x^2-4x+5=4\\\\2x^2-4x+5-4=0\\\\2x^2-4x+1=0\\\\a=2,\ \ b=-4,\ \ c=1\\\\\Delta =b^2-4ac=(-4)^2-4*2*1=16-8=8\\\\\sqrt{\Delta }=\sqrt{8}=\sqrt{4*2}=2\sqrt{2}\\\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-(-4)-2\sqrt{2}}{2*2}=\frac{4-2\sqrt{2}}{2*2}= \frac{2(2-\sqrt{2})}{2*2}= \frac{ 2-\sqrt{2} }{ 2}\\\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-(-4)+2\sqrt{2}}{2*2}=\frac{4+2\sqrt{2}}{2*2}= \frac{2(2+\sqrt{2})}{2*2}= \frac{ 2+\sqrt{2} }{ 2}[/tex]
