\(\left(\sqrt{15}-10\sqrt{11}\right)-\left(\sqrt{20}-6\sqrt{11}\right)=\sqrt{15}-10\sqrt{11}-\sqrt{20}+6\sqrt{11}=\sqrt{15}-4\sqrt{11}-2\sqrt{5}\)
\(\left(\sqrt{15}-10\sqrt{11}\right)-\left(\sqrt{20}-6\sqrt{11}\right)=\sqrt{15}-10\sqrt{11}-\sqrt{20}+6\sqrt{11}=\sqrt{15}-4\sqrt{11}-2\sqrt{5}\)
Đưa về hằng đẳng thức các biểu thức sau:
a) 19+8√3
b)11-4√6
c)9-4√2
d)21+6√10
e)23+6√10
f)49-20√6
Viết các biểu thức sau dưới dạng bình phương của 1 tổng hay 1 hiệu
1/ 7-2√6
2/10 + 2√21
3/11 +2√33
4/10 + 4√16
5/11 - 2√30
6/11 + 2√10
7/11 - 4√6
8/ 11 + 4√7
9/12 - 4√5
10/ 12 + 6√3
\(C=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
\(B=\dfrac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D= \(\dfrac{\left(5+4\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D=\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Mình sửa lại để m.n dễ nhìn hơn!
So sánh A = 2\(\sqrt{1}+2\sqrt{3}+2\sqrt{5}+2\sqrt{7}+2\sqrt{9}+2\sqrt{11}+2\sqrt{13}+2\sqrt{15}+2\sqrt{17}+2\sqrt{19}\) và B = \(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+2\sqrt{8}+2\sqrt{10}+2\sqrt{12}+2\sqrt{14}+2\sqrt{16}+2\sqrt{18}+2\sqrt{20}\)
các bn giúp mk giải bài một nha
giải phương trình 4(x\(^2\)+4x+2) =11\(\sqrt{x^4+4}\)