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5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
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a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}}{3}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2-\sqrt{5}-2\)
\(=-\sqrt{5}\)
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1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
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c: Ở hai hàm số trên, nếu lấy biến x cùng một giá trị thì f(x) sẽ nhỏ hơn g(x) 3 đơn vị
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a) \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
b) \(\left(1+2\sqrt{3}-\sqrt{2}\right)\left(1+2\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+2\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+4\sqrt{3}+12-2\)
\(=9+4\sqrt{3}\)
1/2 + 1/2 = 1 nhé
âm căn 3 + căn 3 hết .