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konrad509
[tex]\frac{1}{x-1}+\frac{1}{x+2}=\frac{5}{4}|\cdot4(x-1)(x+2)\\ D:x\not=1 \wedge x\not=-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[/tex]
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