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![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(A=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
Câu 2:
\(\Leftrightarrow\left|2x-3\right|=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2\sqrt{3}\\2x-3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}+3}{2}\\x=\dfrac{-2\sqrt{3}+3}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x=\sqrt{\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}}>0\)
\(x^2=\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}+2\sqrt{\dfrac{25-24}{25-6}}=\dfrac{74}{19}+\dfrac{2\sqrt{19}}{19}\)
\(\Rightarrow x^2=\dfrac{74+2\sqrt{19}}{19}\Rightarrow x=\sqrt{\dfrac{74+2\sqrt{19}}{19}}\)
Ko thể rút gọn thêm nữa (có thể trục căn thức ở mẫu)
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\sqrt{\frac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\frac{\left(5-2\sqrt{6}\right)^2}{25-24}}=5+2\sqrt{6}+5-2\sqrt{6}=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(E=1+1=2\)
b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)
\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)
d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
=\(\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
= \(\frac{\sqrt{3}+3+\sqrt{2}-\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{2}+\sqrt{5}\right)}\)
= \(\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{5}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}}\)
= \(\sqrt{3}+\sqrt{5}+3\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{2.3}+2}+\sqrt{3-2\sqrt{2.3}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)
\(=2\sqrt{3}\) ( Thiếu bước giá trị tuyệt đối vì minh lười :)) )