Ta có \(\sqrt{17-12\sqrt{2}}=\sqrt{8-12\sqrt{2}+9}\)=\(\sqrt{\left(2\sqrt{2}\right)^2-2.3.2\sqrt{2}+3^2}=\sqrt{\left(2\sqrt{2}-3\right)^2}\)
=/\(2\sqrt{2}-3\)/=\(3-2\sqrt{2}\)(vì 3=\(\sqrt{9},2\sqrt{2}=\sqrt{8}\)=> 3-\(2\sqrt{2}>0\))
rút gọn
\(\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Tính \(\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{17}-12\sqrt{2}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(\sqrt{\left(\sqrt{7}-5\right)^2}+\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17}-12\sqrt{2}}\)
Tính
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
cần gấp
Tính giá trị của biểu thức:
a)A=\(\sqrt{\left(2-\sqrt{5}\right)^2}\) +\(\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
b)B=\(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
c)C=\(\sqrt{17+12\sqrt{2}}\) + \(\sqrt{17-12\sqrt{2}}\)
Giúp vs, làm câu nào cx đc, làm hết thì tốt
a) \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
b) \(\sqrt{27-10\sqrt{2}}+\sqrt{18-8\sqrt{2}}\)
c) \(\sqrt{3-\sqrt{5}}.\sqrt{8}\)
d) \(\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}.\sqrt{8}\)
e) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
g) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-\left(\sqrt{2}+3\right)\)
h) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
i) \(\left(\sqrt[3]{25}-\sqrt[3]{10}+\sqrt[3]{4}\right).\left(\sqrt[3]{5}+\sqrt[3]{2}\right)\)
k) \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
L) \(A=\sqrt[3]{10+14\sqrt{2}}+\sqrt[3]{10-14\sqrt{2}}\)
