Tính: 1) √(52-16√3) +√(〖(4√3-7)〗^2 ) 2) 1/(2+√3)-2/(3+√3)+√2/√6 3) √(14-6√5) – √(29-12√5) 29/08/2021 Bởi Eloise Tính: 1) √(52-16√3) +√(〖(4√3-7)〗^2 ) 2) 1/(2+√3)-2/(3+√3)+√2/√6 3) √(14-6√5) – √(29-12√5)
a. $\sqrt{52 – 16\sqrt{3}} + \sqrt{(4\sqrt{3} – 7)^2} =$$= \sqrt{(4\sqrt{3})^2 – 2.4\sqrt{3}.2 + 2^2} + |4\sqrt{3} – 7| =$$= \sqrt{(4\sqrt{3} – 2)^2} + 7 – 4\sqrt{3}$$= 4\sqrt{3} – 2 + 7 – 4\sqrt{3} = 5$ b. $\dfrac{1}{2 + \sqrt{3}} – \dfrac{2}{3 + \sqrt{3}} + \dfrac{\sqrt{2}}{\sqrt{6}}$ $= \dfrac{2 – \sqrt{3}}{(2 + \sqrt{3})(2 – \sqrt{3})} – \dfrac{2(3 – \sqrt{3})}{(3 + \sqrt{3})(3 – \sqrt{3})} + \dfrac{\sqrt{3}}{3}$$= \dfrac{2 – \sqrt{3}}{1} – \dfrac{6 – 2\sqrt{3}}{9 – 3} + \dfrac{1}{3}\sqrt{3}$$= 2 – \sqrt{3} – 1 + \dfrac{1}{3}\sqrt{3} + \dfrac{1}{3}\sqrt{3}$ $= 1 – \dfrac{1}{3}.\sqrt{3}$c. $\sqrt{14 – 6\sqrt{5}} – \sqrt{29 – 12\sqrt{5}} =$$= \sqrt{(3 – \sqrt{5})^2} – \sqrt{(2\sqrt{5} – 3)^2}$ $= |3 – \sqrt{5}| – |2\sqrt{5} – 3|$ $= 3 – \sqrt{5} – 2\sqrt{5} + 3 = 6 – \sqrt{5}$ Bình luận
a. $\sqrt{52 – 16\sqrt{3}} + \sqrt{(4\sqrt{3} – 7)^2} =$
$= \sqrt{(4\sqrt{3})^2 – 2.4\sqrt{3}.2 + 2^2} + |4\sqrt{3} – 7| =$
$= \sqrt{(4\sqrt{3} – 2)^2} + 7 – 4\sqrt{3}$
$= 4\sqrt{3} – 2 + 7 – 4\sqrt{3} = 5$
b. $\dfrac{1}{2 + \sqrt{3}} – \dfrac{2}{3 + \sqrt{3}} + \dfrac{\sqrt{2}}{\sqrt{6}}$
$= \dfrac{2 – \sqrt{3}}{(2 + \sqrt{3})(2 – \sqrt{3})} – \dfrac{2(3 – \sqrt{3})}{(3 + \sqrt{3})(3 – \sqrt{3})} + \dfrac{\sqrt{3}}{3}$
$= \dfrac{2 – \sqrt{3}}{1} – \dfrac{6 – 2\sqrt{3}}{9 – 3} + \dfrac{1}{3}\sqrt{3}$
$= 2 – \sqrt{3} – 1 + \dfrac{1}{3}\sqrt{3} + \dfrac{1}{3}\sqrt{3}$ $= 1 – \dfrac{1}{3}.\sqrt{3}$
c. $\sqrt{14 – 6\sqrt{5}} – \sqrt{29 – 12\sqrt{5}} =$
$= \sqrt{(3 – \sqrt{5})^2} – \sqrt{(2\sqrt{5} – 3)^2}$
$= |3 – \sqrt{5}| – |2\sqrt{5} – 3|$
$= 3 – \sqrt{5} – 2\sqrt{5} + 3 = 6 – \sqrt{5}$
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