a: \(=\dfrac{2\sqrt{7}+10-2\sqrt{7}+10}{7-25}=\dfrac{-20}{18}=\dfrac{-10}{9}\)
b: \(=\dfrac{7+10\sqrt{7}+25+7-10\sqrt{7}+25}{-18}\)
\(=\dfrac{64}{-18}=\dfrac{-32}{9}\)
rút gọn các biểu thức sau:
\(\dfrac{1}{2}\sqrt{20}+5\)
\(\sqrt{16}+\sqrt{64}\)
\(\sqrt{20}-\sqrt{45}+3\sqrt{18}\)
\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}\)
A= \(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\)
B=\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
mình cần gấp á. tại vì mình khá là ngu toán nên giúp mik vs
Chứng tỏ giá trị các biểu thức sau là số hữu tỉ
a) \(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\)
b) \(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(\dfrac{\sqrt{\dfrac{5}{3}}}{\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1}+\dfrac{\sqrt{\dfrac{7}{5}}}{\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{\sqrt{\dfrac{3}{7}}}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{3}{7}}+1}\)
tính
Bài : Thu gọn
1) \(\dfrac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}\)
2) \(\dfrac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)
3) \(\dfrac{7+4\sqrt{3}}{2+\sqrt{3}}\)
4) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
5) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
6) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6-2\sqrt{10}}}\)
tính giá trị biểu thức :
a, \(\left(\sqrt{28}+2\sqrt{14}+\sqrt{7}\right)\sqrt{7}-\left(7+\sqrt{2}\right)^2\)
b, \(\sqrt{\dfrac{5}{2}}+\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{5}}\)
c, \(\dfrac{8+2\sqrt{15}}{\sqrt{5}+\sqrt{3}}+\dfrac{7-2\sqrt{10}}{\sqrt{5}-\sqrt{2}}\)
d,\(\dfrac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
1. A=\(\sqrt{4+\sqrt{7}}\) +\(\sqrt{4-\sqrt{7}}\)
2. B= \(\dfrac{\sqrt{\sqrt{7-\sqrt{3}}-\sqrt{7+\sqrt{3}}}}{\sqrt{7-\sqrt{2}}}\)
3. C=\(\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
4. D=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
5 E=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\) +\(\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
6. so sánh Cho A=\(\sqrt{11+\sqrt{96}}\)
B= \(\dfrac{2\sqrt{2}}{1+\sqrt{2-\sqrt{3}}}\) so sánh A và b
\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\):\(\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn căn thức sau:
\(B= \dfrac{\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
\(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
