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\(\frac{x^2+3x-1}{2-x}+x>0\Leftrightarrow\frac{5x-1}{2-x}>0\Rightarrow\frac{1}{5}< x< 2\)
\(\frac{\left(x-1\right)^3\left(x+2\right)^2\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x=-2\\1\le x< 2\\2< x< 7\end{matrix}\right.\)
Kết hợp lại ta có: \(1\le x< 2\)
ĐKXĐ: \(x\ge2\)
Khi đó ta có \(x^2-x+1\ge3\Rightarrow1-2\sqrt{x^2-x+1}< 0\)
Do đó BPT tương đương:
\(\sqrt{2\left(x^2+7x+3\right)}-\sqrt{x^2+x-6}-3\sqrt{x+1}\le0\)
\(\Leftrightarrow\sqrt{2x^2+14x+6}\le\sqrt{x^2+x-6}+3\sqrt{x+1}\)
\(\Leftrightarrow2x^2+14x+6\le x^2+10x+3+6\sqrt{\left(x+1\right)\left(x^2+x-6\right)}\)
\(\Leftrightarrow x^2+4x+3\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le6\sqrt{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}\le6\sqrt{x-2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\le36\left(x-2\right)\)
\(\Leftrightarrow x^2-32x+75\le0\)
\(\Rightarrow16-\sqrt{181}\le x\le16+\sqrt{181}\)
ĐKXĐ: \(\left[{}\begin{matrix}x>3\\x\le-1\end{matrix}\right.\)
- Với \(x>3\) BPT tương đương:
\(\left(x-3\right)\left(x+1\right)+2\sqrt{\left(x-3\right)\left(x+1\right)}-3< 0\)
\(\Leftrightarrow\left(\sqrt{\left(x-3\right)\left(x+1\right)}-1\right)\left(\sqrt{\left(x-3\right)\left(x+1\right)}+3\right)< 0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 1\)
\(\Leftrightarrow x^2-2x-4< 0\Rightarrow3< x< 1+\sqrt{5}\)
- Với \(x\le-1\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)-2\sqrt{\left(x-3\right)\left(x+1\right)}< 3\)
\(\Leftrightarrow\left(\sqrt{\left(x-3\right)\left(x+1\right)}+1\right)\left(\sqrt{\left(x-3\right)\left(x+1\right)}-3\right)< 0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 9\Leftrightarrow x^2-2x-12< 0\)
\(\Rightarrow1-\sqrt{13}< x\le-1\)
Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}3< x< 1+\sqrt{5}\\1-\sqrt{13}< x\le-1\end{matrix}\right.\)