T = \(\dfrac{\sqrt{5}\left(\sqrt{16}-\sqrt{9}\right)}{4-5}-5\sqrt{5}+\dfrac{1}{\sqrt{5}-2}+2\sqrt{5}\)

   = \(-\sqrt{5}-5\sqrt{5}+2\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)

   = \(-4\sqrt{5}+\dfrac{1}{\sqrt{5}-2}\)

   = \(\dfrac{-4\sqrt{5}\left(\sqrt{5}-2\right)+1}{\sqrt{5}-2}\)

   = \(\dfrac{-20+8\sqrt{5}+1}{\sqrt{5}-2}\)

   = \(\dfrac{-19+8\sqrt{5}}{\sqrt{5}-2}\)

   = \(\dfrac{19-8\sqrt{5}}{2-\sqrt{5}}\)

   = \(\dfrac{\left(-2+3\sqrt{5}\right)\left(\sqrt{5}-2\right)}{-\left(\sqrt{5}-2\right)}=2-3\sqrt{5}\)

Ta có: \(T=\dfrac{\sqrt{80}-\sqrt{45}}{4-\sqrt{25}}-\sqrt{125}+\dfrac{1}{\sqrt{5}-2}+\dfrac{2\sqrt{55}}{\sqrt{11}}\)

\(=\dfrac{4\sqrt{5}-3\sqrt{5}}{-1}-5\sqrt{5}+\sqrt{5}+2+2\sqrt{5}\)

\(=3\sqrt{5}-4\sqrt{5}-5\sqrt{5}+\sqrt{5}+2+2\sqrt{5}\)

\(=-3\sqrt{5}+2\)

\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)

9.

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)

\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=-\sqrt{5}\)

10.

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)

\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)

\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)

\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\\ =2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}=\sqrt{5}\)

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\ =5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\ =\dfrac{15\sqrt{3}-4\sqrt{3}+6\sqrt{6}+18\sqrt{3}}{3}\\ =\dfrac{29\sqrt{3}+6\sqrt{6}}{3}\)

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)

\(=5\sqrt{5}-1\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

a) \(\sqrt{ }\)20 + 2\(\sqrt{ }\)45 - 3\(\sqrt{ }\)80 + \(\sqrt{ }\)125

= \(\sqrt{ }\)4.5 +2\(\sqrt{ }\)9.5 - 3\(\sqrt{16.5}\)

= 2\(\sqrt{5}\) + 6\(\sqrt{5}\) - 12\(\sqrt{5}\)

= -4\(\sqrt{5}\)

b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) - \(4\sqrt{\dfrac{3}{2}}\)- \(\dfrac{5}{1-\sqrt{6}}\)

= \(\dfrac{2\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)- \(\sqrt{16.\dfrac{3}{2}}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{\left(1-\sqrt{6}\right)\left(1+\sqrt{6}\right)}\)

= 2 - \(\sqrt{24}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{1-6}\)

= 2 - \(\sqrt{4.6}\) + 1+\(\sqrt{ }\)6

= 2 - 2\(\sqrt{ }\)6 + 1+\(\sqrt{ }\)6

= 3 - \(\sqrt{ }\)6

c) (đề bài) với x khác 4...

= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)- \(\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

= \(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)- ....

= \(x-4\sqrt{x}+4\)/ \(\sqrt{x}\left(\sqrt{x}-2\right)\)

= (căn -2)²/ căn x(căn x -2)

= căn x-2/căn x

a) căn 5

Bài 1:

a) \(\sqrt{20}+2\sqrt{45}-3\sqrt{80}+\sqrt{125}\)

  = \(2\sqrt{5}+6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\)

  = \(\sqrt{5}\)

b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}-4\sqrt{\dfrac{3}{2}}-\dfrac{5}{1-\sqrt{6}}\)

  = \(\left(2\sqrt{3}+3\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)-2\sqrt{6}+1+\sqrt{6}\)

  = \(\sqrt{6}-\sqrt{6}+1=1\)

c) \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\left(\sqrt{x}-1\right)}{x-2\sqrt{x}}\) (ĐKXĐ: x > 0; x ≠ 4)

  = \(\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}\)

  = \(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

  = \(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

\(=\sqrt{3}+1-2+\sqrt{3}\)

\(=2\sqrt{3}-1\)

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+2\)

=3

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

\(=\sqrt{3}+1-2+\sqrt{3}\)

\(=2\sqrt{3}-1\)

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+2\)

=3

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

Mysterious Person giúp e với! Em cảm ơn!!!

a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)

c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)

e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)