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a, Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định thì (x-1)(x-3)\(\ge\)0
TH1: \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\Leftrightarrow}x\ge3}\)TH2:\(\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\Leftrightarrow}x\le1}\) Vậy nếu \(x\ge3\) hoặc \(x\le1\) thì biểu thức có nghĩa
b, Để \(\sqrt{x^2-4}=\sqrt{\left(x-2\right)\left(x+2\right)}\)có nghĩa thì (x-2)(x+2)\(\ge0\)
TH1: \(\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge-2\end{matrix}\right.\Leftrightarrow x\ge}2}\)TH2:\(\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\le-2\end{matrix}\right.\Leftrightarrow}x\le-2}\)Vậy nếu \(x\ge2\) hoặc \(x\le-2\) thì biểu thức có nghĩa
a/ \(P=\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}=\dfrac{\sqrt{x}-x+x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)
\(=\dfrac{\sqrt{x}-x+x\sqrt{x}+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)\(=\dfrac{\sqrt{x}+x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)}{\left(\sqrt{x}+1\right)\sqrt{x}\left(1-\sqrt{x}\right)}\)\(=\dfrac{x+1}{1-x}\)
b/ thay x = \(\dfrac{1}{\sqrt{2}}\) vào P:
\(P=\dfrac{\dfrac{1}{\sqrt{2}}+1}{1-\dfrac{1}{\sqrt{2}}}=3+2\sqrt{2}\)
Bài 2:
a: ĐKXĐ: 2/3x-1/5>=0
=>2/3x>=1/5
hay x>=3/10
b: ĐKXĐ: \(\dfrac{x+1}{2x-3}>=0\)
=>2x-3>0 hoặc x+1<=0
=>x>3/2 hoặc x<=-1
c: ĐKXĐ: \(\left\{{}\begin{matrix}3x-5>=0\\x-4>0\end{matrix}\right.\Leftrightarrow x>4\)
a) bt xác định
<=> x^2-4x+3>=0
<=> x^2-4x+4-1>=0
<=> (x-2)^2-1>=0
<=> (x-2)^2>=1
<=> x-2>=1 hoặc x-2<=1
Đến đây bạn giải 2 trường hợp trên là ra kết quả
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
a: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{3x+\sqrt{x}+3-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\cdot\left(3-2\sqrt{2}\right)+2\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1}\)
\(=\dfrac{6-4\sqrt{2}+2\sqrt{2}-2+2}{\sqrt{2}-1}=\dfrac{6-2\sqrt{2}}{\sqrt{2}-1}=4\sqrt{2}+2\)
\(ĐKXĐ:\left\{{}\begin{matrix}\dfrac{x^2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne-1\end{matrix}\right.\Leftrightarrow x>-1\)