Tính:
\(H=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
Tính: \(H=\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(H=\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)\(-\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)(cái này cùng dòng với cái phía trên)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{2\sqrt{3}}\)
\(H=\frac{-4}{2\sqrt{3}}\)
\(H=\frac{-2}{\sqrt{3}}\)
\(H=-\frac{2\sqrt{3}}{3}\)
Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(A^2=4+2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(A^2=4+2=6\)
\(A=\sqrt{6}\)
Đặt \(B=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(B^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(B^2=4-2\sqrt{1}=4-2=2\)
\(B=\sqrt{2}\)
Thay vào H
\(\Rightarrow H=\frac{\sqrt{2}}{\sqrt{6}}-\frac{\sqrt{6}}{\sqrt{2}}=\frac{1}{\sqrt{3}}-\sqrt{3}=\frac{1-3}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\)
Giúp mình với, ko cần làm hết đâu. Tính!
a)\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b)\(\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
c)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
d)\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Tính: a)\(\frac{3\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}+\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\)
b) \(6\sqrt{\frac{1}{3}}-\frac{\sqrt{3}-3}{\sqrt{3}}+\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{2}-\sqrt{3}}\)
c) \(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-2\sqrt{2}\)
d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Xin giúp 4 bài trên. Cảm ơn trước ạ!
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
Thực hiện phép tính:
a)\(\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
b)\(\sqrt{\frac{3-2\sqrt{2}}{7-12\sqrt{2}}}-\sqrt{\frac{3+2\sqrt{2}}{17+12\sqrt{2}}}\)
c)\(\sqrt{6-\sqrt{6-\sqrt{25-\sqrt{96}}}}\)
d)\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
e)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\)
f)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
g)\(\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
h) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}\)
Tính:
a, \(\frac{2}{5+2\sqrt{6}}+\frac{20}{\sqrt{6}-1}+\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}-\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b,\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
b/ \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n+1}-\sqrt{n}\)
\(=\sqrt{n+1}-1\)
Câu a quy đồng từ từ từ phải qua trái là ra
Tính giá trị biểu thức
A = \(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
B = \(\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
\(x=\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}}}}}}}}}}{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+\sqrt{\frac{2}{\sqrt{3+1}}}}}}}}}}}}}}}\)
Tính \(A=\left(\sqrt{x}^{1000}+x^{500}\right)^{2000}\)
Dễ thấy x có tử = 2; mẫu = 1. Vậy x = 2.
\(A=\left(2^{500}+2^{500}\right)^{2000}=2^{501.2000}\)
eo ơi Mr Lazy nhìn sao ra tớ nhìn ko hỉu nỗi
a. P=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}+\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
b.P= (\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}\)) ( 5+\(\sqrt{27}\))
c. P= (\(\frac{2+\sqrt{2}}{\sqrt{2}+1}+1\))(\(\frac{2-\sqrt{2}}{\sqrt{2}-1}-1\))
d. P=\(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}\)
đ. P=(2+\(\sqrt{4+\sqrt{6+2\sqrt{5}}}\) )(\(\sqrt{10}-\sqrt{2}\) )
e. P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
ê. P= \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
g. G= \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
h. H=\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
i. I= \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
a)
\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3+1-2\sqrt{3.1}}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{3}-1}{\sqrt{2}(\sqrt{3}-1)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9-2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}-3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}-3)}\)
\(=\sqrt{9}=3\)
\(\Rightarrow P=\frac{\sqrt{2}}{2}+3\)
b)
\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}-\frac{52(3\sqrt{3}+1)}{(3\sqrt{3}-1)(3\sqrt{3}+1)}+\frac{12(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}\)
\(=\frac{2(\sqrt{3}+1)}{2}-\frac{52(3\sqrt{3}+1)}{26}+\frac{12(3+\sqrt{3})}{6}\)
\(=\sqrt{3}+1-2(3\sqrt{3}+1)+2(3+\sqrt{3})=9\sqrt{3}+9=5-3\sqrt{3}\)
\(\Rightarrow P=(5-3\sqrt{3})(5+3\sqrt{3})=-2\)
c)
\(P=\left[\frac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1}+1\right]\left[\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}-1\right]\)
\(=(\sqrt{2}+1)(\sqrt{2}-1)=2-1=1\)
d)
\(P\sqrt{2}=\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}-2=\sqrt{17+1+2\sqrt{17.1}}-\sqrt{17+1-2\sqrt{17.1}}-2\)
\(=\sqrt{(\sqrt{17}+1)^2}-\sqrt{(\sqrt{17}-1)^2}-2=(\sqrt{17}+1)-(\sqrt{17}-1)-2=0\)
\(\Rightarrow P=0\)
đ)
\(2+\sqrt{4+\sqrt{6+2\sqrt{5}}}=2+\sqrt{4+\sqrt{5+1+2\sqrt{5.1}}}=2+\sqrt{4+\sqrt{(\sqrt{5}+1)^2}}\)
\(=2+\sqrt{4+\sqrt{5}+1}=2+\sqrt{5+\sqrt{5}}\)
\(\Rightarrow P=(2+\sqrt{5+\sqrt{5}})(\sqrt{10}-\sqrt{2})\), cái số này rút gọn không có ý nghĩa, sẽ ra số rất xấu, bạn xem lại đề.
e)
\(P=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1+\sqrt{2})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Tính
H=\(\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}}{2-\sqrt{6}}+\frac{\sqrt{3}}{2+\sqrt{6}}\right)-\frac{1}{\sqrt{2}}\)