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a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
a. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+14\sqrt{2}=14-14\sqrt{2}+7+14\sqrt{2}=21\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}-\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
c. \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
MK CHỈ VIẾT KQ THUI NHA ! VÌ DÀI QUÁ ...
A= 4,236067977 B = 2,414213562 C= 0,8218544151
D= 3,968118785 E= \(-\)\(10\sqrt{2}\) F=17,10050878 (\(3\sqrt{5}+6\sqrt{3}\))
G=\(-7\sqrt{5}\) H= \(-10\sqrt{2}\)
K VÀ KB NHOA ! 
Dương Nguyễn Ngọc Khánh !
Bài 3:
a) Ta có: \(4+2\sqrt{3}\)
\(=3+2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}+1\right)^2\)
b) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
c) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)
d) Ta có: \(31+10\sqrt{6}\)
\(=25+2\cdot5\cdot\sqrt{6}+6\)
\(=\left(5+\sqrt{6}\right)^2\)
e) Ta có: \(13+4\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot1+1\)
\(=\left(2\sqrt{3}+1\right)^2\)
g) Ta có: \(21+12\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot3+9\)
\(=\left(2\sqrt{3}+3\right)^2\)
h) Ta có: \(29+12\sqrt{5}\)
\(=20+2\cdot2\sqrt{5}\cdot3+3\)
\(=\left(2\sqrt{5}+3\right)^2\)
i) Ta có: \(49+8\sqrt{3}\)
\(=48+2\cdot4\sqrt{3}\cdot1\)
\(=\left(4\sqrt{3}+1\right)^2\)
k) Sửa đề: \(14-6\sqrt{5}\)
Ta có: \(14-6\sqrt{5}\)
\(=9-2\cdot3\cdot\sqrt{5}+5\)
\(=\left(3-\sqrt{5}\right)^2\)
l) Ta có: \(23-8\sqrt{7}\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=\left(4-\sqrt{7}\right)^2\)
m) Ta có: \(15-4\sqrt{11}\)
\(=11-2\cdot\sqrt{11}\cdot2+4\)
\(=\left(\sqrt{11}-2\right)^2\)
n) Sửa đề: \(28-10\sqrt{3}\)
Ta có: \(28-10\sqrt{3}\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=\left(5-\sqrt{3}\right)^2\)
o) Ta có: \(17-12\sqrt{2}\)
\(=9-2\cdot3\cdot2\sqrt{2}+8\)
\(=\left(3-2\sqrt{2}\right)^2\)
p) Ta có: \(43-30\sqrt{2}\)
\(=25-2\cdot5\cdot3\sqrt{2}+18\)
\(=\left(5-3\sqrt{2}\right)^2\)
q) Ta có: \(51-10\sqrt{2}\)
\(=50-2\cdot5\sqrt{2}\cdot1\)
\(=\left(5\sqrt{2}-1\right)^2\)
r) Ta có: \(49-12\sqrt{5}\)
\(=45-2\cdot3\sqrt{5}\cdot2+4\)
\(=\left(3\sqrt{5}-2\right)^2\)
c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)
g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)
h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
Câu e đâu bạn?