Câu hỏi của Hiếu Nguyễn

Question

1) Tínha) $\dfrac{1}{2-\sqrt{3}}$-3$\sqrt{\dfrac{1}{3}}$$+\sqrt{12}$b) $\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}$

YangSu · Accepted Answer

$a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\
=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\
=\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\
=2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\
=2+2\sqrt{3}$$b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\
=\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\
=\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\
=-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\
=-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\
=-2+2\sqrt{2}-2\sqrt{2}+1\
=-1$Câu trả lời: $a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\
=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\
=\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\
=2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\
=2+2\sqrt{3}$$b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\
=\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\
=\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\
=-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\
=-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\
=-2+2\sqrt{2}-2\sqrt{2}+1\
=-1$

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