\(\sqrt{10-2\sqrt{21}}\)= \(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)=/ \(\sqrt{7}\)+ \(\sqrt{3}\)/ (giá trị tuyệt đối /)= \(\sqrt{7}\)+ \(\sqrt{3}\) ( do \(\sqrt{7}\)+\(\sqrt{3}\) >0)
=> \(\sqrt{a}\)+ \(\sqrt{b}\)= \(\sqrt{7}\)+ \(\sqrt{3}\)
=> a+b= 7+3=10
nếu \(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\) thì a - b = ?
nếu \(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\)
thì a-b=?
Nếu \(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\)với \(a,b\in Z\) thì a - b = ?
\(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\) tính a-b= ?
\(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\)
Tìm hiệu a-b.
Neu \(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b}\) va a,b thuoc Z thi a-b =?
\(\sqrt{10-2\sqrt{21}}=\sqrt{a}-\sqrt{b},vớia,b\in Z,thìa-b=\)
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, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
Rút gọn : ( giúp với )
a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}\)
