Bài 1:

a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)

b) Ta có: \(x^2+2x+1\)

\(=\left(x+1\right)^2\)

mà \(\left(x+1\right)^2\ge0\forall x\)

nên \(x^2+2x+1\ge0\forall x\)

Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x

c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)

\(\Leftrightarrow x\left(x-4\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)

Bài 3:

a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)

\(=\left|3-\sqrt{10}\right|\)

\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))

b) Ta có: \(\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|\sqrt{5}-2\right|\)

\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))

c) Ta có: \(3x-\sqrt{x^2-2x+1}\)

\(=3x-\sqrt{\left(x-1\right)^2}\)

\(=3x-\left|x-1\right|\)

\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)

1) \(\frac{1}{\sqrt{2x-1}}\)có nghĩa khi \(\hept{\begin{cases}2x-1\ge0\\\sqrt{2x-1}\ne0\end{cases}}\)

\(\Leftrightarrow2x-1>0\)

\(\Leftrightarrow x>\frac{1}{2}\)

\(\sqrt{5-x}\)có nghĩa khi \(5-x\ge0\Leftrightarrow x\ge5\)

Vậy \(ĐKXĐ:\frac{1}{2}>x\ge5\)

2) \(\sqrt{x-\frac{1}{x}}\)có nghĩa khi \(\hept{\begin{cases}x-\frac{1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x^2}{x}-\frac{1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x^2-1}{x}\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-1\ge0\\x>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2\ge1\\x>0\end{cases}}\)

Vậy \(ĐKXĐ:x\ge1\)

3) \(\sqrt{2x-1}\)có nghĩa khi \(2x-1\ge0\) \(\Leftrightarrow x\ge\frac{1}{2}\)

\(\sqrt{4-x^2}\)có nghĩa khi \(4-x^2\ge0\Leftrightarrow x^2\le4\Leftrightarrow x\le2\)

Vậy \(ĐKXĐ:\frac{1}{2}\le x\le2\)

4) \(\sqrt{x^2-1}\)có nghĩa khi \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow x\ge1\)

\(\sqrt{9-x^2}\)có nghĩa khi \(9-x^2\ge0\Leftrightarrow x^2\le9\Leftrightarrow x\le3\)

Vậy \(ĐKXĐ:1\le x\le3\)

Bài 1:

1. \(\sqrt{a}\)có nghĩa <=> \(a\ge0\)

2. a) \(\sqrt{2x+6}\)có nghĩa <=> \(2x+6\ge0\)

\(\Leftrightarrow2x\ge-6\)

\(x\ge-3\)

b)\(\sqrt{\frac{-2}{2x-3}}\) có nghĩa \(\Leftrightarrow\frac{-2}{2x-3}\ge0\)

có -2 < 0

\(\Leftrightarrow\hept{\begin{cases}2x-3\ne0\\2x-3\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x\ne3\\2x\le3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne\frac{3}{2}\\x\le\frac{3}{2}\end{cases}}\)

\(\Rightarrow x< \frac{3}{2}\)

Bài 4 :

\(P=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right).\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)

\(\Leftrightarrow\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right):\left(\frac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right):\left(\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow\left(\frac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\right)\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}\) \(\left(ĐKXĐ:x>0;x\ne4;x\ne1\right)\)

b) \(P=\frac{1}{4}\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\)

\(\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}-3\sqrt{x}=8\)

\(\Leftrightarrow\sqrt{x}=8\)

\(\Leftrightarrow x=64\left(TMĐXĐ\right)\)

Vậy khi \(P=\frac{1}{4}\) thì x=64

Bài 4 : 

c) Xét \(x=4+2\sqrt{3}\)(tmđkxđ)

\(=3+2\sqrt{3}+1\)

\(=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Thay \(x=\sqrt{3}+1\)vào biểu thức P ta có :

\(\frac{\left(\sqrt{3}+1\right)-2}{3.\left(\sqrt{3}+1\right)}=\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{3.\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}=\frac{\left(\sqrt{3}-1\right)^2}{3.\left(3-1\right)}=\frac{2-\sqrt{3}}{3}\)

Vậy khi \(x=4+2\sqrt{3}\)thì P = \(\frac{2-\sqrt{3}}{3}\)

d) \(P\in Z\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}\in Z\)

\(\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{\sqrt{x}}{3\sqrt{x}}-\frac{2}{3\sqrt{x}}=\frac{1}{3}-\frac{2}{3\sqrt{x}}\)

\(\Rightarrow\frac{2}{3\sqrt{x}}\in Z\Leftrightarrow2⋮3\sqrt{x}\Leftrightarrow3\sqrt{x}\inƯ\left(2\right)\)

Mà Ư(2) = \(\left\{\pm2;\pm1\right\}\)

Ta có bảng sau 

Không có giá trị thích hợp của  \(x\in Z\)thỏa mãn điều kiện \(P\in Z\)