Lời giải:
\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\sqrt{7}\right).\frac{1}{\sqrt{7}}=\left(\frac{-3}{\sqrt{7}}+\sqrt{7}\right).\frac{1}{\sqrt{7}}=\frac{-3}{7}+1=\frac{4}{7}\)
1, 1+\(\sqrt{\text{6}+2\sqrt{5}}\)
2, \(\sqrt{\text{7}-2\sqrt{\text{10}}}\) +\(\sqrt{\text{2}}\)
3, \(\sqrt{\text{7}+4\sqrt{3}}\)
1]
a) \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
b) \(\left(\frac{\sqrt{1}}{7}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}\right):\sqrt{7}\)
Tính
a) \(\frac{10}{\sqrt{5}}+\frac{8}{3 +\sqrt{5}}-\frac{\sqrt{15}-2\sqrt{5}}{\sqrt{3}-2}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{5}-\sqrt{7}}\)
bài 1 rút gọn
a \(A=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
b\(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
c\(C=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) d\(D=\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
Tính:
A=\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
B=\(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
C=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D=\(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)(2 cách)
F=\(\dfrac{\sqrt{17-12\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}-\dfrac{\sqrt{17}+12\sqrt{2}}{\sqrt{3+2\sqrt{2}}}\)
a, \(\dfrac{\sqrt[]{7-2\sqrt[]{6}}}{\sqrt[]{6}-1}\)
b, 2.|x+y|.\(\sqrt[]{\dfrac{1}{x^2+2xy+y^2}}\) (x+y>0)
c, \(\dfrac{\left(x-5\right)^4}{\left(4-x\right)^2}\)-\(\dfrac{x^2-25}{x-4}\)(x<4)
\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}\)
Bài tập: Thực hiện phép tính
\(\left(\dfrac{1}{3}\sqrt{\dfrac{1}{2}}-\dfrac{2}{3}\sqrt{\dfrac{1}{3}}+\dfrac{2}{7}\sqrt{\dfrac{4}{5}}\right):\left(\dfrac{2}{7}\sqrt{\dfrac{1}{8}}\right)\)
\(\sqrt{3}×\sqrt{27}-\sqrt{144}:\sqrt{36}\)
\(\left(2\sqrt{9}+3\sqrt{36}\right):4\)
\(\sqrt{7}-\sqrt{8-2\sqrt{7}}\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(\sqrt{27}+5\sqrt{12}-2\sqrt{3}=11\sqrt{3}\)
