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\(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\) đề thế này phải k bn
\(\dfrac{\sqrt{2}\left(\sqrt{\sqrt{21}+5}+\sqrt{5-\sqrt{21}}\right)}{\sqrt{2}}\)
=\(\dfrac{\sqrt{3+7+2\sqrt{3.7}}+\sqrt{3+7-2\sqrt{21}}}{\sqrt{2}}\)
=\(\dfrac{\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
=\(\dfrac{\sqrt{3}+\sqrt{7}+\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
=\(\sqrt{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2 :
a) \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+2\sqrt{7}+1}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\left|\sqrt{7}+1\right|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
b) \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{4+4\sqrt{3}+3}-2\sqrt{3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=\left|2+\sqrt{3}\right|-2\sqrt{3}\)
\(=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
c) \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)
\(=\sqrt{13-2\sqrt{13}+1}+\sqrt{13+2\sqrt{13}+1}\)
\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)
\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)
\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)
d) \(D=\sqrt{22-2\sqrt{21}}+\sqrt{22+2\sqrt{21}}\)
\(=\sqrt{21-2\sqrt{21}+1}+\sqrt{21+2\sqrt{21}+1}\)
\(=\sqrt{\left(\sqrt{21}-1\right)^2}+\sqrt{\left(\sqrt{21}+1\right)^2}\)
\(=\left|\sqrt{21}-1\right|+\left|\sqrt{21}+1\right|\)
\(=\sqrt{21}-1+\sqrt{21}+1=2\sqrt{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{32}+10\sqrt{7}+\sqrt{32}-10\sqrt{7}\)
\(=\left(\sqrt{32}+\sqrt{32}\right)+\left(10\sqrt{7}-10\sqrt{7}\right)\)
\(=\sqrt{16\times2}\times2\)
\(=\sqrt{\left(4\right)^2\times2}\times2\)
\(=4\sqrt{2}\times2\)
\(=8\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\)
=\(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge6\)(1)
mặt khác 5-2x-x2=6-(x+1)2\(\le6\)(2)
từ (1) và (2)=>dấu = xảy ra khi VP =6 =VTtức x=-1
b)\(\sqrt{3x^2+6x+12}\)+\(\sqrt{5x^4+10x^2+9}\)
=\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2+1\right)^2+4}>5\)(x2+1>0)(1')
mặt khác VP=5-2(x+1)2\(\le\)5(2')
từ (1') và (2')=> pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{5+\sqrt{21}}=\frac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{14}+\sqrt{6}}{2}\)