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BA
24 tháng 11 2019
\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)
\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2019
Lời giải:
a)
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3+1-2\sqrt{3}}\)
\(=\sqrt{(3+1+2\sqrt{3})+2+(2\sqrt{6}+2\sqrt{2})}-\sqrt{(\sqrt{3}-\sqrt{1})^2}\)
\(=\sqrt{(\sqrt{3}+1)^2+2\sqrt{2}(\sqrt{3}+1)+2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{(\sqrt{3}+1+\sqrt{2})^2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{3}+1+\sqrt{2}-(\sqrt{3}-1)=2+\sqrt{2}\)
b)
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)
\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
AH
Akai Haruma
Giáo viên
17 tháng 9 2019
Lời giải:
a)
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3+1-2\sqrt{3}}\)
\(=\sqrt{(3+1+2\sqrt{3})+2+(2\sqrt{6}+2\sqrt{2})}-\sqrt{(\sqrt{3}-\sqrt{1})^2}\)
\(=\sqrt{(\sqrt{3}+1)^2+2\sqrt{2}(\sqrt{3}+1)+2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{(\sqrt{3}+1+\sqrt{2})^2}-\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{3}+1+\sqrt{2}-(\sqrt{3}-1)=2+\sqrt{2}\)
b)
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)
\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
23 tháng 6 2018
\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}=\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
\(=\sqrt{2+3+1+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}=\left|\sqrt{2}+\sqrt{3}+1\right|=\sqrt{2}+\sqrt{3}+1\)
NP
0
DQ
1
MD
18 tháng 6 2017
Ta có: A = \(\sqrt{\left(\sqrt{6}-2\sqrt{2}\right)^2}-\sqrt{24-12\sqrt{3}}\)
= \(\left|\sqrt{6}-2\sqrt{2}\right|\) \(-\sqrt{18-2.6\sqrt{3}+6}\)
= \(2\sqrt{2}-\sqrt{6}-\sqrt{\left(\sqrt{18}-\sqrt{6}\right)^2}\)
= \(2\sqrt{2}-\sqrt{6}-\sqrt{18}+\sqrt{6}\)
= \(2\sqrt{2}-3\sqrt{2}=-\sqrt{2}\)
\(A=\frac{\sqrt{6+\sqrt{12}-\sqrt{8}-\sqrt{24}}}{\sqrt{2}+\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+1^2-2.\sqrt{2}.\sqrt{3}+2.\sqrt{3}-2.\sqrt{2}}}{\sqrt{2}+\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{2}-\sqrt{3}-1\right)^2}}{\sqrt{2}+\sqrt{3}+1}\)
\(=\frac{\left|\sqrt{2}-\sqrt{3}-1\right|}{\sqrt{2}+\sqrt{3}+1}\)
\(=\frac{1+\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}+1}\)