Bài 2 : Rút gọn biểu thức sau A = sqrt(5 - 2sqrt(6)) - sqrt((sqrt(2) - sqrt(3)) ^ 2)
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12 tháng 6 2023
\(\sqrt{9-3\sqrt{8}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}+\sqrt{5-2\sqrt{6}}-\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{6}+\sqrt{2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{6}-\sqrt{3}\right|-\sqrt{6}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}+\sqrt{3}-\sqrt{2}\) (do \(\sqrt{6}-\sqrt{3}>0;\sqrt{3}-\sqrt{2}>0\))
\(=0\)
12 tháng 6 2023
\(=\sqrt{9-6\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{3}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}\)
\(=0\)
DV
19 tháng 6 2021
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
\(A=\sqrt{5-2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}\)
=0