1) \(\sqrt{\dfrac{1}{200}}\)                            2) \(\dfrac{5}{1-\sqrt{6}}\)

\(=\sqrt{\dfrac{1^2}{10^2.2}}\)                          \(=\dfrac{1-\sqrt{6}+4+\sqrt{6}}{1-\sqrt{6}}\)

\(=\dfrac{1}{10\sqrt{2}}\)                              \(=1+\dfrac{4+\sqrt{6}}{1-\sqrt{6}}\)

Bài 2: 

1. \(\sqrt{2x-5}=7\)    ĐKXĐ: \(x\ge\dfrac{5}{2}\)

<=> 2x - 5 = 7²

<=> 2x - 5 = 49

<=> 2x = 54

<=> x = 27 (TM)

2. \(3+\sqrt{x-2}=4\)     ĐKXĐ: \(x\ge2\)

<=> \(\sqrt{x-2}=1\)

<=> x - 2 = 1

<=> x = 3 (TM)

3. \(\sqrt{x^2-2x+1}=1\)

<=> \(\sqrt{\left(x-1\right)^2}=1\)

<=> \(|x-1|=1\)

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

<=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

4. \(\sqrt{x^2-4x+4}=1\)

<=> \(\sqrt{\left(x-2\right)^2}=1\)

<=> \(|x-2|=1\)

<=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

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

5. \(\sqrt{4x^2+1-4x}=\sqrt{x^2+16+8x}\)

<=> \(\left(\sqrt{4x^2+1-4x}\right)^2=\left(\sqrt{x^2+16+8x}\right)^2\)

<=> \(|4x^2+1-4x|=|x^2+16+8x|\)

<=> \(\left[{}\begin{matrix}4x^2+1-4x=x^2+16+8x\\4x^2+1-4x=-\left(x^2+16+8x\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x^2-x^2-4x-8x+1-16=0\\4x^2+1-4x=-x^2-16-8x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}3x^2-12x-15=0\\5x^2+4x+17=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}3x^2+3x-15x-15=0\\VNghiệm\end{matrix}\right.\)

<=> 3x(x + 1) - 15(x + 1) = 0

<=> (3x - 15)(x + 1) = 0

<=> \(\left[{}\begin{matrix}3x-15=0\\x+1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

1 turning the radio down

2 not interrupting me all the time

3 looking after the baby for you

4 to post these letter

5 asking so many questions

6 smoking 3 years ago

7 having made so many mistakes

8 having bought this book

9 not going to university

10 meeting me

\(A1+2=A2\)

   \(\overline{A}\) =\(\dfrac{A1\cdot54+\cdot\left(A1+2\right)\cdot46}{100}\)=79.92

\(\Leftrightarrow\)A1=79\(\Rightarrow\)A2=81  

Thanksss bann nhiều lắm nhoa

1: \(\sqrt{\dfrac{1}{200}}=\dfrac{\sqrt{2}}{20}\)

2: \(\dfrac{5}{1-\sqrt{6}}=-1-\sqrt{6}\)

3: \(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{1+\sqrt{2}}\)

\(=\dfrac{1+\sqrt{2}-1+\sqrt{2}}{-1}\)

\(=-2\sqrt{2}\)