\(\frac{1}{x-1}+\frac{1}{x+2}=\frac{5}{4}|\cdot4(x-1)(x+2)\\
D:x
ot=1 \wedge x
ot=-2\\
4(x+2)+4(x-1)=5(x-1)(x+2)\\
4x+8+4x-4=5(x^2+2x-x-2)\\
8x+4=5x^2+5x-10\\
5x^2-3x-14=0\\
5x^2-10x+7x-14=0\\
5x(x-2)+7(x-2)=0\\
(5x+7)(x-2)=0\\
x=-\frac{7}{5}\vee x=2\)
