\(9-12x+4x^2>0\)
\(\Rightarrow\left(2-2x\right)^2>0\)
\(\Rightarrow2-2x>0\)
\(\Rightarrow-2x>-2\)
\(\Rightarrow x< 1\)
Vậy để A có nghĩa thì \(x< 1\)
B) \(\sqrt{x+2\sqrt{x-1}}\ne0\)
\(x+2\sqrt{x-1}>0\)
\(\Rightarrow x-1+2\sqrt{x-1}+1>0\)
\(\Rightarrow\left(\sqrt{x-1}+1\right)^2>0\)
\(\sqrt{x-1}\ge0\Rightarrow x\ge1\)\(\)
Vậy \(x\ge1\)thì B có nghĩa
C) \(\sqrt{3x-2}.\sqrt{x-1}\ge0\)
\(\orbr{\begin{cases}3x-2\ge0\\x-1\ge0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ge\frac{2}{3}\\x\ge1\end{cases}}\)
Vậy \(x\ge1\)thì C có nghĩa
a) \(\frac{1}{\sqrt{9-12x+4x^2}}=\frac{1}{\sqrt{\left(2x-3\right)^2}}=\frac{1}{2x-3}\)
để căn thức A có nghĩa \(\Rightarrow2x-3\ne0\Leftrightarrow x\ne\frac{3}{2}\)
b)\(\frac{1}{\sqrt{x+2\sqrt{x}+1}}=\frac{1}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{1}{\sqrt{x}+1}\)
để căn thức B có nghĩa => \(\sqrt{x}+1\ne0\) và \(x\ge0\) hay \(\sqrt{x}+1>1\Leftrightarrow x=0\)
Vậy..........
\(A=\frac{1}{\sqrt{9-12x+4x^2}}=\frac{1}{\sqrt{\left(3-2x\right)^2}}=\frac{1}{|3-2x|}.\)
\(đkxđ\Leftrightarrow3-2x\ne0\Rightarrow2x\ne3\Rightarrow x\ne\frac{3}{2}\)
\(B=\frac{1}{\sqrt{x+2\sqrt{x-1}}}\)\(=\frac{1}{\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}\)\(=\frac{1}{|\sqrt{x-1}+1|}\)
\(đkxđ\Leftrightarrow x-1\ge0\Rightarrow x\ge1\)
\(C=\sqrt{3x-2}.\sqrt{x-1}=\sqrt{\left(3x-2\right)\left(x-1\right)}\)
\(đkxđ\Leftrightarrow\left(3x-2\right)\left(x-1\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}3x-2\ge0;x-1\ge0\\3x-2< 0;x-1< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\ge\frac{2}{3};x\ge1\\x< \frac{2}{3};x< 1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\ge1\\x< \frac{2}{3}\end{cases}}\)
\(9-12x+4x^2>0\)
\(=>\left(2-2x\right)^2>0\)
\(=>2-2x>0\)
~Study well~
