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a)\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{2}}\right).\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\left(ĐKXĐ:x\ne1;x\ge0\right)\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}.\left[\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2\right]}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(2\sqrt{x}\right)\left(-2\right)}{x-1}\right]\)
\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{-4x}{x-1}\right]\)
\(=\frac{-\sqrt{2x}\left(\sqrt{2x}-1\right)}{\left(x-1\right)}\)
\(=\frac{\sqrt{2x}-2x}{\left(x-1\right)}\)
1)\(\sqrt{2x^2-2x+\frac{1}{2}}=\frac{1}{\sqrt{2}}\left(ĐKXĐ:x^2-x+\frac{1}{4}\ge0\right)\)
\(2x^2-2x+\frac{1}{2}=\frac{1}{2}\)
\(2x^2-2x=0\)
\(2x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
2)\(\sqrt{9x-9}-2\sqrt{\frac{x-1}{4}}=6\left(ĐKXĐ:x\ge1\right)\)
\(\sqrt{9\left(x-1\right)}-2.\frac{\sqrt{x-1}}{2}=6\)
\(3\sqrt{x-1}-\left(\sqrt{x-1}\right)=6\)
\(2\sqrt{x-1}=6\)
\(\sqrt{x-1}=3=\sqrt{9}\)
\(\Rightarrow x=10\)
4)\(1-3x+\sqrt{x^2-6x+9}=0\)
\(1-3x+\sqrt{\left(x-3\right)^2}=0\)
\(1-3x+x-3=0\)
\(x=-1\)
5)\(\frac{1}{2}\sqrt{\frac{3x+9}{4}}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{1}{2}.\frac{\sqrt{3x+9}}{2}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{\sqrt{3x+9}}{4}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{\sqrt{3x+9}+4\sqrt{x+3}}{4}=\frac{4\sqrt{1-x}}{4}\)
\(\Rightarrow\sqrt{3}.\sqrt{x+3}+4\sqrt{x+3}=4\sqrt{1-x}\)
\(\Rightarrow\left(\sqrt{3}+4\right)\left(\sqrt{x+3}\right)=\sqrt{2-2x}\)
6)\(\sqrt{4x^2-9}.\left(\sqrt{x+1}+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x^2-9=0\\\sqrt{x+1}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x^2=9\\\sqrt{x+1}=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)
a)biểu thức có nghĩa khi :
-x4 -2 > 0 <=> - x4 > 2
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
Lời giải:
a) ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$
b) ĐKXĐ: $3+2x>0\Leftrightarrow x>\frac{-3}{2}$
c) ĐKXĐ: $x^2-4\geq 0\Leftrightarrow (x-2)(x+2)\geq 0$
$\Leftrightarrow x\geq 2$ hoặc $x\leq -2$
d)
ĐKXĐ\(\left\{\begin{matrix} x\geq 0\\ \sqrt{x}\neq 2\\ x+1>0\\ x\neq 0\\ \sqrt{x}\neq 3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>0\\ x\neq 4\\ x\neq 9\end{matrix}\right.\)
e)
ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ 7-\sqrt{x}>0\end{matrix}\right.\Leftrightarrow 0\leq x< 49\)
f)
\(\left\{\begin{matrix} 5-x\neq 0\\ \frac{x+3}{5-x}\geq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x+3\geq 0\\ 5-x>0\end{matrix}\right.\\ \left\{\begin{matrix} x+3\leq 0\\ 5-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow -3\leq x< 5\)
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) \(x^2-9\ge0\Leftrightarrow x^2\ge9\Leftrightarrow\orbr{\begin{cases}x\ge3\\x\ge-3\end{cases}}\)
b) \(-x-2\ge0\Leftrightarrow-x\ge2\Leftrightarrow x\ge-2\)
c) \(x^2+2x+1=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^2\ge0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)