Tính
a/ A=\(\sqrt{6-2\sqrt{5}}\) - \(\sqrt{5}\)
b/ B=\(\sqrt{17\sqrt{2}-24}\)+ 4
c/ C=\(\sqrt{12+6\sqrt{3}-4}\)- \(\sqrt{3}\)
Bài 3: Thực hiện các phép tính sau:
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
c) \(\sqrt{6-4\sqrt{2}}+\)\(\sqrt{22-12\sqrt{2}}\)
hộ mk với
a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=2\sqrt{5}+2+\sqrt{5}-2\)
\(=3\sqrt{5}\)
b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
\(=3-2\sqrt{2}+2\sqrt{2}-1\)
=2
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
thực hiện phép tính
a)\(\sqrt{6+2\sqrt{5}}-\sqrt{62\sqrt{5}}\)
b)\(\sqrt{24-8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
d)\(\sqrt{41+12\sqrt{5}}-\sqrt{46-6\sqrt{ }5}\)
e)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
f)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
g) \(\sqrt{43+24\sqrt{3}}-\sqrt{49-8\sqrt{3}}\)
cho cách làm dạng bài này luôn. Chỗ nào chưa hiểu thì nói tớ sẽ giải thích thêm (cần góp ý để hoàn thiện thêm phần hướng dẫn đó mà. Cảm ơn cậu).
Phương Nam Phim (à quên, Từ Hạ) hân hạnh giới thiệu bộ phim...
Thực hiện phép tính:
a)\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
b)\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
c)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
d)\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)
b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)
c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)
d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)
Bài 1: Chứng minh:
a) \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}=0\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
c) \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
d) \(\sqrt{6+\sqrt{6+\sqrt{6+,,,+\sqrt{6}}}}=3\)
Tính giá trị của biểu thức:
a)A=\(\sqrt{\left(2-\sqrt{5}\right)^2}\) +\(\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
b)B=\(\sqrt{6+2\sqrt{5}}\) - \(\sqrt{6-2\sqrt{5}}\)
c)C=\(\sqrt{17+12\sqrt{2}}\) + \(\sqrt{17-12\sqrt{2}}\)
a) A= \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
Vì \(\left\{{}\begin{matrix}2=\sqrt{4}< \sqrt{5}\\2\sqrt{2}=\sqrt{8}>\sqrt{5}\end{matrix}\right.\) nên A = \(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
= \(\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\)
= \(2\left(\sqrt{2}-1\right)\)
b) B = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\) (B > 0)
Ta có:
B2 = \(6+2\sqrt{5}-2\sqrt{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}+6-2\sqrt{5}\)
= \(12-2\sqrt{36-20}\)
= \(12-8\)
= \(4\)
\(\Rightarrow\) B =\(\pm2\) nhưng vì B > 0 nên B = 2
Vậy B = 2
c) C = \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\) (C > 0)
Ta có:
C2 = \(17+12\sqrt{2}+2\sqrt{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}+\left(17-12\sqrt{2}\right)\)
= \(34+2\sqrt{289-288}\)
= \(34+2\)
= \(36\)
\(\Rightarrow C=\pm6\) nhưng vì C > 0 nên C = 6
Tính
a/ A=\(\sqrt{6-2\sqrt{5}}\) - \(\sqrt{5}\)
b/ B=\(\sqrt{17\sqrt{2}-24}\)+ 4
c/ C=\(\sqrt{12+6\sqrt{3}-4}\)- \(\sqrt{3}\)
a)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{5}\\ A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\\ A=\sqrt{5}-1-\sqrt{5}\\ A=-1\)
Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
Bài 1: Tính
A=\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
B=\(\sqrt{13-\sqrt{160}-\sqrt{53+4\sqrt{90}}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
D=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
E= \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
F= \(\sqrt{3+\sqrt{11+6\sqrt{2}}}-\sqrt{5+2\sqrt{6}}\)
G=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Bài 2: so sánh
a) \(\sqrt{24}+\sqrt{45}\) và 12
b) \(\sqrt{37}-\sqrt{15}\) và 2
c) \(\sqrt{16}\) và \(\sqrt{15}\times\sqrt{17}\)
d) 8 và \(\sqrt{15}+\sqrt{17}\)
Bài 2 :
a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)
b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)
c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)
Tính P=\(a^3+b^3-3\left(a+b\right)+2012\)
Biết \(a=\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}};b=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)