Rut gon:
\(B=\frac{\sqrt{8+\sqrt{40+8\sqrt{5}}}+\sqrt{8-\sqrt{40+8\sqrt{5}}}}{\sqrt{50}+\sqrt{250}}\)
Những câu hỏi liên quan
\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
RUT GON
rut gon:
a)\(3\sqrt{8}-4\sqrt{18}+2\sqrt{50}\)
b)\(5\sqrt{12}+2\sqrt{75}-5\sqrt{48}\)
c)\(\frac{a}{b}\sqrt{\frac{b}{a}}-\frac{1}{a}\sqrt{a^3b}+\frac{2}{3b}\sqrt{9ab^3}\left(a,b>0\right).\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) rut gon bieu thuc gium em a thanks
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Đúng 0
Bình luận (0)
\(A=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\frac{8}{x}+\frac{16}{^{x^2}}}}\)
rut gon
Rut gon dum minh
\(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
Mong co cau tra loi som nhat
Rút gọn
a) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
b)\(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{5}}+2\sqrt{5}-\sqrt{3}\)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+\sqrt{2.4}+\sqrt{5.4}+\sqrt{10.4}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}\right)^2+\left(\sqrt{2}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{2}.\sqrt{1}+2\sqrt{1}.\sqrt{5}+2\sqrt{5}.\sqrt{2}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}+\sqrt{2}+\sqrt{5}\right)^2}\)
= \(\sqrt{1}+\sqrt{2}+\sqrt{5}\)
Đúng 0
Bình luận (4)
CMR A = \(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{2\sqrt{3}+2\sqrt{5}}+....+\frac{1}{40\sqrt{79}+40\sqrt{81}}\) <\(\frac{8}{9}\)
Bài 1: Tính
\(\sqrt{3+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\\ \sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}\\ \sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}\\ \sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}\\ \sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\\ \sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
1.
$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3}}-\sqrt{3+1-2\sqrt{3}}$
$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}$
$=|\sqrt{3}+1|-|\sqrt{3}-1|=2$
2.
\(\sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}=\sqrt{12+6\sqrt{3}+\sqrt{9+3-2\sqrt{9.3}}}=\sqrt{12+6\sqrt{3}+\sqrt{(3-\sqrt{3})^2}}\)
\(=\sqrt{12+6\sqrt{3}+3-\sqrt{3}}=\sqrt{15+5\sqrt{3}}\)
Đúng 0
Bình luận (0)
3.
\(\sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{9-4\sqrt{2}+\sqrt{8+1+2\sqrt{8.1}}}\)
\(=\sqrt{9-4\sqrt{2}+\sqrt{2\sqrt{2}+1)^2}}=\sqrt{9-4\sqrt{2}+2\sqrt{2}+1}=\sqrt{10-2\sqrt{2}}\)
4.
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8+1-2\sqrt{8.1}}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{(\sqrt{8}-1)^2}}}\) \(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8}-1}}=\sqrt{\sqrt{2}+2+\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{(2+1+2\sqrt{2}}}=\sqrt{\sqrt{2}+2+\sqrt{(\sqrt{2}+1)^2}}=\sqrt{\sqrt{2}+2+\sqrt{2}+1}\)
\(=\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1\)
Đúng 0
Bình luận (0)
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
rut gon bt ;
a) \(\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
b) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
c) \(9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)
d) \(\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(a,\left(\sqrt{27}-2\sqrt{17}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=3\sqrt{21}-2\sqrt{119}+7+7\sqrt{8}\)
Đề sai chăng???
Đúng 0
Bình luận (0)
\(b,\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}-1+\sqrt{2}+1\)
\(=2\sqrt{2}\)
\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)
\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)
\(=\sqrt{2}\left(9-8-5+8\right)\)
\(=4\sqrt{2}\)
\(d,\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4+2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2-\sqrt{2}\)
\(=-3\)
Đúng 0
Bình luận (0)
\(c,9\sqrt{2}-4\sqrt{8}-\sqrt{50}+2\sqrt{32}\)
\(=9\sqrt{2}-8\sqrt{2}-5\sqrt{2}+8\sqrt{2}\)
\(=4\sqrt{2}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời