Question

17 va rog , dau coroana.

Answer 1

$\displaystyle \it 17a). \frac{x+1}{x+2} +2 \cdot \left(1+ \frac{1-x}{x-2} \right)= \frac{2x+1}{x+2} - \frac{(x-3)(x+4)}{x^2-4} \\ \\ \frac{x+1}{x+2} +2 \cdot \frac{x-2+1-x}{x-2} = \frac{2x+1}{x+2} - \frac{x^2+4x-3x-12}{x^2-4} \\ \\ \frac{x+1}{x+2} +2 \cdot \frac{-1}{x-2} = \frac{2x+1}{x+2} - \frac{x^2+x-12}{x^2-4} \\ \\ \frac{x+1}{x+2} - \frac{2}{x-2} = \frac{2x+1}{x+2} - \frac{x^2+x-12}{(x-2)(x+2)}$
$\displaystyle \it (x+1)(x-2)-2(x+2)=(2x+1)(x-2)-(x^2+x-12) \\ \\ x^2-2x+x-2-2x-4=2x^2-4x+x-2-x^2-x+12 \\ \\ x^2-2x^2+x^2-2x+x-2x+4x-x+x=-2+12+2+4 \\ \\ x=16$
$\displaystyle \it b). \frac{x-2}{x+1}- \frac{3}{x} =2 \left( \frac{1}{2x} - \frac{x}{x+1} \right)+ \frac{(3x-1)(x-2)}{x^2+x} \\ \\ \frac{x-2}{x+1} - \frac{3}{x} = 2 \cdot \frac{x+1-2x \cdot x}{2x(x+1)} + \frac{3x^2-6x-x+2}{x(x+1)} \\ \\ \frac{x-2}{x+1} - \frac{3}{x} =2 \cdot \frac{x+1-2x^2}{2x(x+1)} + \frac{3x^2-7x+2}{x(x+1)}$
$\displaystyle \it \frac{x(x-2)-3(x+1)}{x(x+1)} = \frac{2x+2-4x^2}{2x(x+1)} + \frac{3x^2-7x+2}{x(x+1)} \\ \\ \frac{x^2-2x-3x-3}{x(x+1)} = \frac{2x+2-4x^2}{2x(x+1)} + \frac{3x^2-7x+2}{x(x+1)} \\ \\ \frac{x^2-5x-3}{x(x+1)} = \frac{2x+2-4x^2}{2x(x+1)} + \frac{3x^2-7x+2}{x(x+1)} \\ \\ 2(x^2-5x-3)=2x+2-4x^2+2(3x^2-7x+2) \\ \\ 2x^2-10x-6=2x+2-4x^2+6x^2-14x+4 \\ \\ 2x^2+4x^2-6x^2-10x-2x+14x=2+4+6 \\ \\ 2x=12 \Rightarrow x= \frac{12}{2} \Rightarrow x=6$
$\displaystyle \it c). \frac{2x+3}{x(x+1)(x+2)} =2 \left( \frac{1}{x} - \frac{1}{x+1} \right)\right+3\left( \frac{1}{x+1}- \frac{1}{x+2} \right) \\ \\ \frac{2x+3}{x(x+1)(x+2)}=2 \cdot \frac{x+1-x}{x(x+1)} +3 \cdot \frac{x+2-(x+1)}{(x+1)(x+2)} \\ \\ \frac{2x+3}{x(x+1)(x+2)}=2 \cdot \frac{1}{x(x+1)} +3 \cdot \frac{x+2-x-1}{(x+1)(x+2)} \\ \\ \frac{2x+3}{x(x+1)(x+2)}= \frac{2}{x(x+1)} + 3 \cdot \frac{1}{(x+1)(x+2)}$
$\displaystyle \it \frac{2x+3}{x(x+1)(x+2)}= \frac{2}{x(x+1)} + \frac{3}{(x+1)(x+2)} \\ \\ 2x+3=2(x+2)+3x \\ \\ 2x+3=2x+4+3x \\ \\ 2x-2x-3x=4-3 \\ \\ -3x=1 \Rightarrow x=- \frac{1}{3}$
$\displaystyle \it d). \frac{x+2}{x-3} + \frac{2x-5}{x+1} =(x+2)\left( \frac{2}{x-3}- \frac{1}{x+1} \right)+ \frac{(2x+1)(x-2)}{x^2-2x-3} \\ \\ \frac{x+2}{x-3} + \frac{2x-5}{x+1} =(x+2) \cdot \frac{2(x+1)-(x-3)}{(x+1)(x-3)} + \frac{2x^2-4x+x-2}{(x+1)(x-3)} \\ \\ \frac{x+2}{x-3} + \frac{2x-5}{x+1}=(x+2) \cdot \frac{2x+2-x+3}{(x+1)(x-3)} + \frac{2x^2-3x-2}{(x+1)(x-3)}$
$\displaystyle \it \\ \\ \frac{x+2}{x-3} + \frac{2x-5}{x+1}=(x+2) \cdot \frac{x+5}{(x-1)(x-3)} + \frac{2x^2-3x-2}{(x+1)(x-3)} \\ \\ \frac{(x+2)(x+1)+(2x-5)(x-3)}{(x+1)(x-3)} = \frac{(x+2)(x+5)}{(x-1)(x-3)} + \frac{2x^2-3x-2}{(x+1)(x-3)} \\ \\ (x+2)(x+1)+(2x-5)(x-3)=(x+2)(x+5)+2x^2-3x-2 \\ \\ x^2+x+2x+2+(2x-5)(x-3)=(x+2)(x+5)+2x^2-3x-2 \\ \\ x^2+3x+2+2x^2-6x-5x+15=(x+2)(x+5)+2x^2-3x-2 \\ \\ 3x^2-8x+17=x^2+5x+2x+10+2x^2-3x-2 \\ \\ 3x^2-x^2-2x^2-8x-5x-2x+3x=10-2-17$
$\displaystyle \it -12x=-9 \Rightarrow x= \frac{9}{12} \Rightarrow x= \frac{3}{4}$