Rozwiązane

Rozwiąż równanie:
1/x+1/(x+1)=(x^2-2)/(x^2+x)



Odpowiedź :

Konrad509

[tex]D:x\not=0 \wedge x+1\not=0 \wedge x^2+x\not=0\\D:x\not=0 \wedge x\not=-1 \wedge x(x+1)\not=0\\D:x\not=0 \wedge x\not=-1 \\\\\\\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{x^2-2}{x^2+x}\Big|\cdot (x^2+x)\\\\x+1+x=x^2-2\\x^2-2x-3=0\\x^2+x-3x-3=0\\x(x+1)-3(x+1)=0\\(x-3)(x+1)=0\\x=3 \vee x=-1\\\\-1\not \in D\\\\\boxed{x=3}[/tex]

Catta1eya

[tex]\frac1x+\frac1{x+1}=\frac{x^2-2}{x^2+x}\\\frac{x+1}{x(x+1)}+\frac{x}{x(x+1)}\\\frac{2x+1}{x^2+x}=\frac{x^2-2}{x^2+x} /* x^2+x \\2x+1=x^2-2\\x^2+x \neq 0\\x\neq 0\\x\neq -1\\D\in R /\{0, -1\}\\\\-x^2+2x+1+2=0\\-x^2+2x+3=0\\\Delta=2^2-4*(-1)*3\\\Delta=4+12\\\Delta=16\\\sqrt{\Delta}=4\\x_1=\frac{-2-4}{-2}=\frac{-6}{-2}=3\\x_2=\frac{-2+4}{-2}=\frac{2}{-2}=-1 - \text{Wykluczone przez dziedzine}\\\\x=3[/tex]

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