rút gọn
\(E=\frac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}+\frac{3+2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\frac{1}{\sqrt{12+2\sqrt{35}}}\)
11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)
Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
a. P=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}+\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
b.P= (\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}\)) ( 5+\(\sqrt{27}\))
c. P= (\(\frac{2+\sqrt{2}}{\sqrt{2}+1}+1\))(\(\frac{2-\sqrt{2}}{\sqrt{2}-1}-1\))
d. P=\(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}\)
đ. P=(2+\(\sqrt{4+\sqrt{6+2\sqrt{5}}}\) )(\(\sqrt{10}-\sqrt{2}\) )
e. P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
ê. P= \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
g. G= \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
h. H=\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
i. I= \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
a)
\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3+1-2\sqrt{3.1}}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{3}-1}{\sqrt{2}(\sqrt{3}-1)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9-2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}-3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}-3)}\)
\(=\sqrt{9}=3\)
\(\Rightarrow P=\frac{\sqrt{2}}{2}+3\)
b)
\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}-\frac{52(3\sqrt{3}+1)}{(3\sqrt{3}-1)(3\sqrt{3}+1)}+\frac{12(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}\)
\(=\frac{2(\sqrt{3}+1)}{2}-\frac{52(3\sqrt{3}+1)}{26}+\frac{12(3+\sqrt{3})}{6}\)
\(=\sqrt{3}+1-2(3\sqrt{3}+1)+2(3+\sqrt{3})=9\sqrt{3}+9=5-3\sqrt{3}\)
\(\Rightarrow P=(5-3\sqrt{3})(5+3\sqrt{3})=-2\)
c)
\(P=\left[\frac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1}+1\right]\left[\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}-1\right]\)
\(=(\sqrt{2}+1)(\sqrt{2}-1)=2-1=1\)
d)
\(P\sqrt{2}=\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}-2=\sqrt{17+1+2\sqrt{17.1}}-\sqrt{17+1-2\sqrt{17.1}}-2\)
\(=\sqrt{(\sqrt{17}+1)^2}-\sqrt{(\sqrt{17}-1)^2}-2=(\sqrt{17}+1)-(\sqrt{17}-1)-2=0\)
\(\Rightarrow P=0\)
đ)
\(2+\sqrt{4+\sqrt{6+2\sqrt{5}}}=2+\sqrt{4+\sqrt{5+1+2\sqrt{5.1}}}=2+\sqrt{4+\sqrt{(\sqrt{5}+1)^2}}\)
\(=2+\sqrt{4+\sqrt{5}+1}=2+\sqrt{5+\sqrt{5}}\)
\(\Rightarrow P=(2+\sqrt{5+\sqrt{5}})(\sqrt{10}-\sqrt{2})\), cái số này rút gọn không có ý nghĩa, sẽ ra số rất xấu, bạn xem lại đề.
e)
\(P=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1+\sqrt{2})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
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) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
Tính giá trị các biểu thức sau:
a) \(A=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
b) \(A=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
c) \(A=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
c) \(A=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
Tính
a)\(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
b)\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{3}-1}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
c)\(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
a) Ta có: \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\frac{\sqrt{5}}{2}\)
\(=\frac{2\sqrt{5}-\sqrt{5}}{2}=\frac{\sqrt{5}}{2}\)
b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{3}-1}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{\left(2\sqrt{8}-\sqrt{12}\right)\left(\sqrt{3}+1\right)}{3-1}-\frac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}+\sqrt{2}\right)}{30-2}\)
\(=\frac{4\sqrt{6}+4\sqrt{2}-6-2\sqrt{3}}{2}-\frac{5\sqrt{6}+\sqrt{10}+9\sqrt{10}+3\sqrt{6}}{28}\)
\(=\frac{4\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{3}\left(\sqrt{3}+1\right)}{2}-\frac{10\sqrt{10}+8\sqrt{6}}{28}\)
\(=\frac{2\cdot\left(\sqrt{3}+1\right)\left(2\sqrt{2}-\sqrt{3}\right)}{2}-\frac{\sqrt{1000}+\sqrt{384}}{28}\)
\(=2\sqrt{6}-3+2\sqrt{2}-\sqrt{3}-\frac{\sqrt{2}\cdot\left(5\sqrt{5}+4\sqrt{3}\right)}{14}\)
\(=\frac{28\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}-8\sqrt{6}}{14}\)
\(=\frac{20\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}}{14}\)
c) Ta có: \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{2}\)
\(=\frac{4}{2}=2\)
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
