Lời giải chi tiết:
ĐKXĐ: \(x\ge-3\)
\(\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)(bình phương 2 vế)
\(\Leftrightarrow x=1\)(TMĐK)
Vậy x=1
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\(\sqrt{72}:\sqrt{8}+\sqrt{3}.\sqrt{12}-\sqrt{27}\)\(=6\sqrt{2}:2\sqrt{2}+\sqrt{36}-3\sqrt{3}\) \(=3+6-3\sqrt{3}\)
\(=9-3\sqrt{3}\)

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Bài 2
a) Để P có nghĩa
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
vậy để P có nghĩ thì \(x\ge0;x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{-2\sqrt{x}}{\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}\right).\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{x-1}.\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(=\dfrac{-2\sqrt{x}}{x-1}.\dfrac{\sqrt{x}-1}{-2\sqrt{x}}\)
\(\dfrac{1}{\sqrt{x}+1}\)


 

a) M= \(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
=\(\left(\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}+\dfrac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\)
=\(\left(\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a-4\sqrt{a}}{a-1}\right)\left(\dfrac{a-1}{\sqrt{a}}\right)\)
=\(\dfrac{4a}{a-1}.\dfrac{a-1}{\sqrt{a}}\)
=\(4\sqrt{a}\)
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???????

\(\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}-\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{2-1}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2 
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*Lời giải chi tiết:
\(10\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)
=\(10\left(12-12\sqrt{6}+18\right)+2\sqrt{6}+6\sqrt{6}\)
=\(120-120\sqrt{6}+180+2\sqrt{6}+6\sqrt{6}\)
=\(300-112\sqrt{6}\)

Tam giác ABC vuông tại A, Ta có
TanB = CotC = 4/3 (2 góc phụ nhau)
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E = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
= \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\sqrt{180}+9}}}\)
=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)}}}\)
=\(\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
=\(\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)
=\(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
=\(\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1 
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