1) 

a) \(\sqrt{2x-4}\) có nghĩa khi:

\(2x-4\ge0\)

\(\Leftrightarrow2x\ge4\)

\(\Leftrightarrow x\ge\dfrac{4}{2}\)

\(\Leftrightarrow x\ge2\)

b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi 

\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)

\(\Rightarrow4-x\le0\)

\(\Leftrightarrow x\ge4\)

2) 

a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)

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

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

\(A=\sqrt{5}+2-\sqrt{5}+2\)

\(A=4\)

\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)

\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=-\left(7-5\right)\)

\(B=-2\)

a) ĐKXĐ: \(-x-8\ge0\Leftrightarrow x\le-8\)

b) ĐKXĐ: \(x^2-2x+1>0\Leftrightarrow\left(x-1\right)^2>0\Leftrightarrow x\ne1\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\5-x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

d) ĐKXĐ: \(x^2+3\ge0\left(đúng.do.x^2+3\ge3>0\right)\)

cứu mị :<

\(a,P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

       \(=\left(1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

       \(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

       \(=\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\left(\frac{3-\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)     

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

       \(=\frac{3}{2-\sqrt{x}}\)

b, Để P > 0 thì \(2-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)(Thỏa mãn DKXD)

\(c,Q=P\left(x+1\right)=\frac{3\left(x+1\right)}{2-\sqrt{x}}\)

Ko biết e đã học miền giá trị chưa nhỉ ???

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 \(P\in Z\)