Question

A=(x+1)/(x-2)+(x-1)/(x+2)+(x^2+3)/(4-x^2).tim x de A ko am

 

Answer 1

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Answer 2

ĐKXĐ: \(x\ne\pm2\)

Ta có : \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+3}{4-x^2}\)

\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2-3x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+3x+2+x^2-3x+2-x^2-3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+1}{x^2-4}\)

Vì \(x^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(\Rightarrow\)Để A không âm thì \(x^2-4>0\)(do \(x\ne\pm2\)nên \(x^2-4\ne0\))

\(\Leftrightarrow x^2>4\)

\(\Leftrightarrow\orbr{\begin{cases}x>2\\x< -2\end{cases}}\)

Vậy để A không âm thì \(x>2\)hoặc \(x< -2\)