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\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
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\(\dfrac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}.\sqrt{5}+2\sqrt{5}-3\sqrt{2}.\sqrt{3}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{\sqrt{5}\left(3\sqrt{2}+2\right)-\sqrt{3}\left(3\sqrt{2}+2\right)}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(3\sqrt{2}+2\right)}{\sqrt{5}+\sqrt{3}}\)
Bạn coi lại xem dưới mẫu đúng dấu ''+'' không á, phải dấu ''-'' mới rút với tử ở trên được nha.
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\(A=\frac{\sqrt{2}-1}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)\(\Leftrightarrow A=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(\Leftrightarrow A=\sqrt{n}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\)
=0
= 0.8740320489
\(\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5-2\sqrt{5}.1+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{2}}=\frac{\sqrt{10}-\sqrt{2}}{2}\)