Rút gon \(T=3\sqrt{2}+2\sqrt{3}-\sqrt{18}+\sqrt{28-16\sqrt{3}}\)
Những câu hỏi liên quan
Rút gon biểu thức sau: \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\)
Rút gọn các biểu thức sau : a) sqrt{18left(sqrt{2}-sqrt{3}right)^2} - sqrt{54}b) dfrac{a+sqrt{ab}}{sqrt{a}-sqrt{b}} - asqrt{dfrac{1}{a}}c) ( sqrt{28} - 2sqrt{3} +7) sqrt{7} +sqrt{84}
Đọc tiếp
Rút gọn các biểu thức sau :
a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) - \(\sqrt{54}\)
b) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\) - a\(\sqrt{\dfrac{1}{a}}\)
c) ( \(\sqrt{28}\) - 2\(\sqrt{3}\) +7) \(\sqrt{7}\) +\(\sqrt{84}\)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
Đúng 1
Bình luận (0)
Rút gon: \(\sqrt{7-2\sqrt{2+\sqrt{50}+\sqrt{18-\sqrt{188}}}}\)
Rút gọn:
a)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
Đúng 0
Bình luận (0)
Rút gọn:
a) \(\frac{\sqrt{6+\sqrt{14}}}{2\sqrt{3+\sqrt{28}}}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Đúng 0
Bình luận (0)
Rút gọn :
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
b) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) = \(1+\sqrt{2}\)
Đúng 0
Bình luận (0)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)
Đúng 0
Bình luận (1)
a) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\) = \(\dfrac{\sqrt{2}}{2}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1. Tìm giá trị nhỏ nhất của M:
M = \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+15+8\sqrt{x-1}}\)
2. Rút gọn:
A= \(\sqrt{\sqrt{28-16\sqrt{3}}}-\sqrt{\sqrt{28+16\sqrt{3}}}\)
B= \(\sqrt{5-2\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
2.
A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)
\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1-\sqrt{3}-1\)
\(=-2\)
B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)
\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{5-2\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)
\(=\sqrt{5-2\sqrt{2}-2}\)
\(=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}\)
\(=\sqrt{2}-1\)
Đúng 0
Bình luận (0)
Rút gọn biểu thức :
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{1}{\sqrt{2}}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
Đúng 0
Bình luận (0)
Rút gọn
a)\(\frac{\sqrt{3}+\sqrt{27}}{2\sqrt{3}+\sqrt{18}}\) b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{2+\sqrt{2}+\sqrt{3}}\)