Rút gọn:\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)
Rút gọn:
\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)
\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)
\(=\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)\)
\(=3^2-\left(\sqrt{3}\right)^2\)
\(=9-3\)
\(=6\)
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn biểu thức \(B=\left(\frac{\sqrt{a-2}+2}{3}\right)\left(\frac{\sqrt{a-2}}{3+\sqrt{a-2}}+\frac{a+7}{11-a}\right):\left(\frac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\frac{1}{\sqrt{a-2}}\right)\)
Rút gọn biểu thức
a)\(\frac{3\sqrt{7}+7\sqrt{3}}{\sqrt{21}}\)
b)\(\frac{\sqrt{2\left(\sqrt{2-\sqrt{7}}\right)^2}}{\sqrt{56}-4}\)
c)\(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right):\left(15-2\sqrt{6}\right)\)
Rút gọn các biểu thức sau
a)\(\left(\sqrt{14}+\sqrt{16}\right)\sqrt{5-\sqrt{21}}\)
b)\(\frac{5}{\sqrt{21}-4}+\frac{3\sqrt{7}-7\sqrt{3}}{\sqrt{7}-\sqrt{3}}\)
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
Rút gọn
1,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
2,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
3,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
4,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
5,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
Sao làm hổng ai bảo đú.n/g vậy :(((
Tính:
a) \(A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
b) \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
c) \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}\)
d) \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
Rút gọn các biểu thức:
\(a,\sqrt{\sqrt{3}+2}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\)
\(b,\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}\)