\(\sqrt{11-2\sqrt{18}}\)= \(\sqrt{\left(3-\sqrt{2}\right)^2}\)= \(3-\)\(\sqrt{2}\)
\(\Rightarrow a=3;b=-1\)\(\Rightarrow ab=-3\)
\(\sqrt{11-2\sqrt{18}}\)= \(\sqrt{\left(3-\sqrt{2}\right)^2}\)= \(3-\)\(\sqrt{2}\)
\(\Rightarrow a=3;b=-1\)\(\Rightarrow ab=-3\)
\(\sqrt{11-2\sqrt{18}}=a+b\sqrt{2}\). Tính ab?
\(\sqrt{11-2\sqrt{18}}=a+b\sqrt{2}\). tính ab=?
a,\(\sqrt{8+2\sqrt{15}}\) -\(\sqrt{6+2\sqrt{15}}\)
b, \(\sqrt{17-2\sqrt{72}}-\sqrt{19+2\sqrt{18}}\)
c, \(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\)
d, \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
e, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
\(\sqrt{11-2\sqrt{18}}\)= a+ b\(\sqrt{2}\)
tinh ab=? ( a, b thuoc Z)
Nếu\(\sqrt{11-2\sqrt{18}}\) = a + b \(\sqrt{2}\), với a,b ∈Z thì a.b = ......
1) Tính: \(\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}+\sqrt{9-2\sqrt{14}}\)
2) Cho a, b là hai số dương thỏa mãn \(\sqrt{ab}=\frac{a+b}{a-b}\)
Tìm GTNN của biểu thức P=\(ab+\frac{a-b}{\sqrt{ab}}\)
Nếu \(\sqrt{11-2\sqrt{18}}=a+b\sqrt{2}\) với \(a,b\in Z\) thì ab = ?
a. \(\dfrac{1}{3}.\sqrt{18}-\sqrt{192}-\dfrac{\sqrt{33}}{\sqrt{11}}+3\sqrt{5\dfrac{1}{3}}\)
b. \(\sqrt{\left(2\sqrt{3}-5\right)^2}-2\sqrt{7+4\sqrt{3}}\)
Cho ab+bc+ca=11.Tìm GTNN của P=\(\frac{5a+5b+2c}{\sqrt{12\left(a^2+11\right)}+\sqrt{12\left(b^2+11\right)}+\sqrt{c^2+11}}\)