Câu hỏi của Nhi Phương

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YangSu · Accepted Answer

$\sqrt{\left(4-3\sqrt{2}\right)^2}=\left|4-3\sqrt{2}\right|=3\sqrt{2}-4$$\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\
\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}$$\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\ \sqrt{7+4\sqrt{3}}=\sqrt{\sqrt{3^2}+2.2\sqrt{3}+2^2}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\
\sqrt{12-6\sqrt{3}}=\sqrt{\sqrt{3^2}-2.3\sqrt{3}+3^2}=\sqrt{\left(\sqrt{3}-3\right)^2}=\left|\sqrt{3}-3\right|=3-\sqrt{3}$$\sqrt{17+12\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.3+3^2}=\sqrt{\left(2\sqrt{2}+3\right)^2}=\left|2\sqrt{2}+3\right|=2\sqrt{2}+3$$\dfrac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}}{\sqrt{\sqrt{5^2}+2\sqrt{5}+1}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}+3\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\left|\sqrt{2}+3\right|}{\left|\sqrt{5}+1\right|-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{2}-3}{\sqrt{5}+1-\sqrt{5}}\
=-3$$\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left|\sqrt{3}-1\right|}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1$Câu trả lời: $\sqrt{\left(4-3\sqrt{2}\right)^2}=\left|4-3\sqrt{2}\right|=3\sqrt{2}-4$$\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\
\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}$$\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\ \sqrt{7+4\sqrt{3}}=\sqrt{\sqrt{3^2}+2.2\sqrt{3}+2^2}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\
\sqrt{12-6\sqrt{3}}=\sqrt{\sqrt{3^2}-2.3\sqrt{3}+3^2}=\sqrt{\left(\sqrt{3}-3\right)^2}=\left|\sqrt{3}-3\right|=3-\sqrt{3}$$\sqrt{17+12\sqrt{2}}=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.3+3^2}=\sqrt{\left(2\sqrt{2}+3\right)^2}=\left|2\sqrt{2}+3\right|=2\sqrt{2}+3$$\dfrac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}}{\sqrt{\sqrt{5^2}+2\sqrt{5}+1}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}+3\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}}\
=\dfrac{\sqrt{2}-\left|\sqrt{2}+3\right|}{\left|\sqrt{5}+1\right|-\sqrt{5}}\
=\dfrac{\sqrt{2}-\sqrt{2}-3}{\sqrt{5}+1-\sqrt{5}}\
=-3$$\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left|\sqrt{3}-1\right|}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1$

Bùi Minh Huy · Answer

-3Câu trả lời: -3

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