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\(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\) đề thế này phải k bn
\(\dfrac{\sqrt{2}\left(\sqrt{\sqrt{21}+5}+\sqrt{5-\sqrt{21}}\right)}{\sqrt{2}}\)
=\(\dfrac{\sqrt{3+7+2\sqrt{3.7}}+\sqrt{3+7-2\sqrt{21}}}{\sqrt{2}}\)
=\(\dfrac{\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
=\(\dfrac{\sqrt{3}+\sqrt{7}+\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
=\(\sqrt{14}\)
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`sqrt{3-sqrt5}-sqrt{3+sqrt5}`
`=sqrt{(6-2sqrt5)/2}-sqrt{(6+2sqrt5)/2}`
`=sqrt{(sqrt5-1)^2/2}-sqrt{(sqrt5+1)^2/2}`
`=(sqrt5-1)/sqrt2-(sqrt5+1)/sqrt2`
`=(sqrt5-1-sqrt5-1)/sqrt2`
`=(-2)/sqrt2=-sqrt2`
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\(\dfrac{1}{\sqrt{5}-2}+\dfrac{\sqrt{10}-\sqrt{5}}{1-\sqrt{2}}=\dfrac{\sqrt{5}+2}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ =\sqrt{5}+2-\sqrt{5}=2\)
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\(\sqrt{\sqrt{5}-\sqrt{5-\sqrt{21-4\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\sqrt{20^2}-2.\sqrt{20}+1}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{\left(\sqrt{20}-1\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\left|\sqrt{20}-1\right|}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-\sqrt{20}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{5^2}-2\sqrt{5}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=1\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
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\(B=50-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=50-3.\sqrt{7^2.2}+2\sqrt{2^2.2}+3\sqrt{4^2.2}-5\sqrt{3^2.2}\)
\(=50-3.7\sqrt{2}+2.2\sqrt{2}+3.4\sqrt{2}-5.3\sqrt{2}\)
\(=50-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)
\(=50+\sqrt{2}.\left(-21+4+12-15\right)\)
\(=50+\sqrt{2}.\left(-20\right)\)
\(=50-20\sqrt{2}\)
\(C=\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{5}-\sqrt{7}\right)\)
\(=\left(\sqrt{3}+\sqrt{5}\right)^2-\sqrt{7}^2\)
\(=\sqrt{3}^2+2.\sqrt{3}.\sqrt{5}+\sqrt{5}^2-7\)
\(=2\sqrt{15}+3+5-7\)
\(=2\sqrt{15}+1\)
Nghĩ ra xong tính thử thấy đúng định nàm xong thấy mẹ giải r ấy:")). Với nại con còn nhỏ nắm, hong bic nhiều cái mà nớp 9 hay sử dụng nữa ý, sợ dùng sai;-;.
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1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)