các bạn giải cho mình bài toán này với so sánh: A=\(2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+.......+2\sqrt{19}\)và B= \(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+....+2\sqrt{18}+\sqrt{20}\)
Cho \(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2.\sqrt{19}\)
và \(B=2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}\)
So sánh A và B
Cho \(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}\)
và \(B=2\sqrt{2}+2\sqrt{4}+2.\sqrt{6}+....+2.\sqrt{18}+\sqrt{20}\)
So sánh A và B
Cho A= \(2\sqrt{1}+2\sqrt{3}+...+2\sqrt{19}\) và B=\(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}\)
So sánh A và B
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
bài 1 : rút gọn
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(B=\frac{3\sqrt{8}-2\sqrt{12}+20}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}-\sqrt{3-\sqrt{5}}\)
các bạn làm hộ mình với ạ , mình đg cần gấp
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
so sánh các số sau:
\(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}\)
\(\text{Và }B=2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{20}\)
Xét hiệu :
\(A-B=2\left(\sqrt{1}-\sqrt{2}\right)+2.\left(\sqrt{3}-\sqrt{4}\right)+...+2\left(\sqrt{19}-\sqrt{20}\right)\)
Mà: \(\sqrt{1}
có ai biết giải bài toán này k giúp mình với ?
1,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}}\)
2,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
3,\(\dfrac{3}{\sqrt{6}-\sqrt{3}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
4,\(\left(\sqrt{\dfrac{2}{3}-\sqrt{\dfrac{3}{2}}+\dfrac{5}{\sqrt{6}}}\right):\dfrac{6-\sqrt{6}}{1-\sqrt{6}}\)
5,\(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\times\left(\sqrt{3}+\sqrt{2}\right)\)
6,\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
7, \(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
8,\(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
9,\(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
10,\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\)
11,\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
12,\(\dfrac{\sqrt{3}+2\sqrt{2}+\sqrt{3}-2\sqrt{2}}{\sqrt{3}+2\sqrt{2}-\sqrt{3}-2\sqrt{2}}\)
1. Sửa đề:
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)
2.
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)
3.
\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\)
\(=\frac{3(\sqrt{6}+\sqrt{3})}{3}+\frac{4(\sqrt{7}-\sqrt{3})}{4}=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)
Bài 1:
a)\(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}\)
b) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{19+2\sqrt{18}}\)
c) \(\sqrt{9+4\sqrt{5}}-\sqrt{\left(1-\sqrt{5}^2\right)}\)
Bài 2: Biến đổi biểu thức
a) \(\dfrac{1}{\sqrt{7}+3}+\dfrac{1}{\sqrt{7}-3}\)
b) \(\dfrac{3}{\sqrt{2}-1}+\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1}\)
c) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\)
So sánh:
\(A=2\sqrt{1}+2\sqrt{3}+...+2\sqrt{19}\)
\(B=2\sqrt{2}+2\sqrt{4}+...+2\sqrt{18}+\sqrt{20}\)