Tính
\(\frac{\sqrt{10}+5\sqrt{3}}{\sqrt{15}+\sqrt{5}}-\frac{3}{2\sqrt{2}-\sqrt{5}}+\sqrt{9+4\sqrt{2}}\)
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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Rút gọn biểu thức
1)\(\frac{15}{3\sqrt{20}}\)
2) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}\)
3) \(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
4) \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
5) \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)
6)\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
7) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
8)\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
9) \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
10) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
11) \(\sqrt{\sqrt{10}+1}.\sqrt{\sqrt{10}-1}\)
12) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
13) \(\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{12}}\)
14) \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\sqrt{6}\)
15 ) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
16) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
17) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
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Thực hiện phép tính
1)\(\frac{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}+\sqrt{2}}{\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}+\sqrt{5}}\)
2)\(\left(4+\sqrt{15}\right)\left(10-\sqrt{6}\right)-\sqrt{4-\sqrt{15}}\)
3)\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
4)\(\frac{2\sqrt{3-\sqrt{5+\sqrt{13-\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
5)\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
Tính:
\(\frac{\sqrt{10}+5\sqrt{3}}{\sqrt{15}+\sqrt{5}}-\frac{3}{2\sqrt{2}-\sqrt{5}}+\sqrt{9+4\sqrt{2}}\)
\(\dfrac{\sqrt{10}+5\sqrt{3}}{\sqrt{15}+\sqrt{5}}-\dfrac{3}{2\sqrt{2}-\sqrt{5}}+\sqrt{9+4\sqrt{2}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{15}\right)}{\sqrt{5}\left(\sqrt{3}+1\right)}-\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{\left(2\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}+\sqrt{8+2.2\sqrt{2}+1}=\dfrac{\sqrt{15}+\sqrt{2}}{\sqrt{3}+1}-\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{8-5}+\sqrt{\left(2\sqrt{2}+1\right)^2}=\dfrac{\left(\sqrt{15}+\sqrt{2}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{3\left(2\sqrt{2}+\sqrt{5}\right)}{3}+2\sqrt{2}+1=\dfrac{3\sqrt{5}-\sqrt{15}+\sqrt{6}-\sqrt{2}}{2}-2\sqrt{2}-\sqrt{5}+2\sqrt{2}+1=\dfrac{3\sqrt{5}-\sqrt{15}+\sqrt{6}-\sqrt{2}}{2}-\sqrt{5}+1=\dfrac{3\sqrt{5}-\sqrt{15}+\sqrt{6}-\sqrt{2}-2\sqrt{5}+2}{2}=\dfrac{\sqrt{5}-\sqrt{15}+\sqrt{6}+2-\sqrt{2}}{2}\)
\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
Bài 2: Thực hiện phép tính
a) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)
c) \(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)
d) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)
f) \(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)
Bài 3: Thực hiện phép tính
a) \(\sqrt{9-4\sqrt{5}}\)
b) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
c) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
d) \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
e) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
f*) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Bài 4: Rút gọn
a) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}\)
b) \(\left(2\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-2\right)\)
c) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
d) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)
e) \(\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)
f) \(\frac{1}{5}\sqrt{50}-2\sqrt{96}-\frac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\frac{1}{6}}\)
Bài 3: Thực hiện phép tính
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\frac{10-2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
3)\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
5)\(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
6)\(6\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
7)\(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
8)\(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
9)\(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)
Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
Trục căn thức ở mẫu và rút gọn:
a) \(\frac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
b) \(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2.\sqrt{3+2\sqrt{5}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)