Rút gọn: A\(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
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C
3
21 tháng 10 2021
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
QA
17 tháng 12 2019
\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{2.\left(2-\sqrt{3}\right)}}{\sqrt{2}}+\frac{\sqrt{2.\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}}\)
\(=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}+\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}+\frac{\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}.\sqrt{2}}{\left(\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{6}}{2}\)
\(=\sqrt{6}\)
Chúc bạn hok tốt!!! The Godlin
ND
25 tháng 7 2018
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{\sqrt{9\left(5+3\sqrt{2}\right)}+\sqrt{9\left(5-3\sqrt{2}\right)}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{3+\sqrt{2}+2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+3-\sqrt{2}}{2\sqrt{2}}\)
\(=\dfrac{3\left[5+3\sqrt{2}+5-3\sqrt{2}+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\right]}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}=\dfrac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}=\dfrac{12}{6\sqrt{2}}\)
\(=\sqrt{2}\)
25 tháng 7 2018
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\sqrt{5+3\sqrt{2}}+3\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)}-\dfrac{\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}}{\sqrt{6+3\sqrt{2}}-\sqrt{6-4\sqrt{2}}}\\ =\dfrac{3\left(5+3\sqrt{2}+2\sqrt{25-18}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{\sqrt{4+2+4\sqrt{2}}+\sqrt{4+2-4\sqrt{2}}}{\sqrt{4+2+4\sqrt{2}}-\sqrt{4+2-4\sqrt{2}}}\\ =\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{2+\sqrt{2}+2-\sqrt{2}}{2+\sqrt{2}-2+\sqrt{2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{4}{2\sqrt{2}}=\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
AH
Akai Haruma
Giáo viên
6 tháng 7 2019
Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
1 tháng 2 2022
\(=\dfrac{a\sqrt{a}-3-2\left(a-6\sqrt{a}+9\right)-a-4\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{a-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-a-4\sqrt{a}-6-2a+12\sqrt{a}-18}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}=\sqrt{a}-1\)
\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A^2=\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)^2\)
\(A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(A^2=4+2\sqrt{4-3}\)
\(A^2=6\)
Vì \(A>0\)\(\Rightarrow A=\sqrt{6}\)
\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\\ A=\frac{\sqrt{2}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\\ A=\frac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\\ A=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\\ A=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\\ A=\frac{2\sqrt{3}}{\sqrt{2}}\\ A=\sqrt{6}\)