\(=2\sqrt{5}-3-\sqrt{20}=-3\)
`\sqrt{(3-2\sqrt{5})^2}-\sqrt{20}`
`=|3-2\sqrt{5}|-\sqrt{2^2 . 5}`
`=2\sqrt{5}-3-2\sqrt{5}` (Vì `2\sqrt{5} > 3`)
`=-3`
\(=2\sqrt{5}-3-\sqrt{20}=-3\)
`\sqrt{(3-2\sqrt{5})^2}-\sqrt{20}`
`=|3-2\sqrt{5}|-\sqrt{2^2 . 5}`
`=2\sqrt{5}-3-2\sqrt{5}` (Vì `2\sqrt{5} > 3`)
`=-3`
thực hiện phép tính
a, \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
b, \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
c, \(\sqrt{2-\sqrt{3}}.\left(\sqrt{5}+\sqrt{2}\right)\)
d, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
e, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
f, \(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\)
g, \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
h, \(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)
i, \(\dfrac{\left(\sqrt{5+2}\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
k, \(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)
l, \(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
m, \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
n, \(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3+1}}}+\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3+1}}}\)
Tính:
a.\(\sqrt{5}-2\sqrt{20}-3\sqrt{80}\)
b.\(\dfrac{6}{\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}\)
c.\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}}{x-1}\) với x > 0;x ≠ 1
Thu gọn biểu thức:
\(B=21\cdot\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
Rút gọn các biểu thức sau:
\(a.A=2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
\(b.B=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(c.C=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}+1}\) (x > 0;x ≠ 4)
Chứng minh rằng:
\(\dfrac{1}{3\left(\sqrt{2}+1\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{1}{7\left(\sqrt{4}+\sqrt{3}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
Cho biểu thức: P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)
a) Rút gọn P.
b) Tìm x để P ≤ -2.
Cho phương trình: P = \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)
a) Rút gọn P.
b) Tìm x để P ≤ -2
a.Thực hiện phép tính:
A = \(-3\sqrt{8}+\sqrt{50}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
b.Rút gọn biểu thức
B = \(\left(\dfrac{5\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-x}\right).\left(1-\dfrac{1}{\sqrt{x}}\right)\) với x > 0 và x≠ 1