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a. Để \(\frac{\sqrt{x-3}}{2x+1}\)có nghĩa thì 2x+1 \(\ne\)0
\(\Leftrightarrow\)2x \(\ne\)-1
\(\Leftrightarrow\)x \(\ne\)\(\frac{-1}{2}\)
b. Để \(\frac{\sqrt{1-2x}}{x^2-6x+9}\) có nghĩa thì x2-6x+9\(\ne\)0
\(\Leftrightarrow\)(x-3)2 \(\ne\)0
\(\Leftrightarrow\)x-3 \(\ne\)0
\(\Leftrightarrow\)x \(\ne\)3
a) ĐKXĐ: \(x^2+6x+11\ge0\)đúng\(\forall x\inℝ\)
b) ĐKXĐ: \(\hept{\begin{cases}\left(2x-3\right)\left(x+2\right)\ge0\\x+3\ne0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\le-2,x\ne-3\\x\ge\frac{3}{2}\end{cases}}}\)
c) ĐKXĐ: \(-x^2-5\ge0\)Vô nghiệm\(\forall x\inℝ\)
a)\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{2}}\right).\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}.\left[\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2\right]}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(2\sqrt{x}\right)\left(-2\right)}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{-4x}{x-1}\right]\)
\(=\frac{-\sqrt{2x}\left(\sqrt{2x}-1\right)}{\left(x-1\right)}\)
\(=\frac{\sqrt{2x}-2x}{\left(x-1\right)}\)
a) \(x^2-9\ge0\Leftrightarrow x^2\ge9\Leftrightarrow\orbr{\begin{cases}x\ge3\\x\ge-3\end{cases}}\)
b) \(-x-2\ge0\Leftrightarrow-x\ge2\Leftrightarrow x\ge-2\)
c) \(x^2+2x+1=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^2\ge0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)