a,A=\(\dfrac{1}{1-\sqrt{x-1}}\)có nghĩa thì 1-(x-1)>0
\(\Leftrightarrow\)-(x-1)> -1
\(\Leftrightarrow\)x-1<1
\(\Leftrightarrow x-1< 1\)
\(\Leftrightarrow\)x<2
b, để \(\dfrac{1}{\sqrt{x^2-2x+1}}\) có nghĩa thì x bình -2x+1>0
\(\Leftrightarrow\)\(\left(x-1\right)^2\)>0
\(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\1-x>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>1\\x< 1\end{matrix}\right.\)
Bài 4:
Ta có: \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\)
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x-6x+5=0\)
\(\Leftrightarrow-11\sqrt{2x}=-11\)
\(\Leftrightarrow2x=1\)
hay \(x=\dfrac{1}{2}\)
Bài 5:
Ta có: \(P=\sqrt{x^2-2x+5}\)
\(=\sqrt{x^2-2x+1+4}\)
\(=\sqrt{\left(x-1\right)^2+4}\ge2\forall x\)
Dấu '=' xảy ra khi x=1
Bài 2:
a) Ta có: \(M=\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}\)
\(=\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)\)
=16-3
=13
b) Ta có: \(N=\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}\)
\(=\dfrac{\sqrt{16-2\sqrt{15}}}{2\left(\sqrt{15}-1\right)}\)
\(=\dfrac{1}{2}\)
Bài 3:
Ta có: \(P=\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)
\(=\left(4+2\sqrt{x}+x+2\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)^2}\)
\(=x-4\sqrt{x}+4\)
