1) \(\dfrac{4\sqrt{x}-x-4}{x-4}=\dfrac{-\left(x-4\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\)

2) \(\dfrac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

3) \(\dfrac{x-9}{x\sqrt{x}-27}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}\right)^3-3^3}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)

hình a

Các vận động viên phải đi trong số ngày là:  ngày

Đáp số:  ngày

1. Although

2. despite

3. although

4. although

5. because of

6. because

7. because

8. because of

9. because of

10. because of 

\(\Leftrightarrow3y-16=12\)

\(\Leftrightarrow3y=28\)

\(\Leftrightarrow y=\dfrac{28}{3}\)

\(\Leftrightarrow3\left(y+7\right)=24\Leftrightarrow y+7=8\Leftrightarrow y=1\)

Vậy, y=1

\(Vì\left\{{}\begin{matrix}\left|x+2\right|\ge0\\\left|x^2+2x\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+2=0\\x\left(x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

Vậy, \(x\in\left\{0;-2\right\}\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

VT= \(\left(\sqrt{14}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right).\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{14}+\sqrt{6}\right).\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right).\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=7-3=4\)