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g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4

15 tháng 7 2017

a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)

b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)

c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)

d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

28 tháng 4 2018

a. \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

= \(\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\)

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

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

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

b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

= \(\dfrac{\left(\sqrt{15}-\sqrt{5}\right).\left(\sqrt{3}+1\right)}{2}+\dfrac{\left(5-2\sqrt{5}\right).\left(2\sqrt{5}+4\right)}{4}\)

=\(\dfrac{\sqrt{45}+\sqrt{15}-\sqrt{15}-\sqrt{5}}{2}+\dfrac{\left(5-2\sqrt{5}\right).2\left(\sqrt{5}+2\right)}{4}\)

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

= \(\dfrac{2\sqrt{5}}{2}+\dfrac{5\sqrt{5}+10-10-4\sqrt{5}}{2}\)

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

= \(\dfrac{3\sqrt{5}}{2}\)

c. \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}\right):\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)

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

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

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

= \(\dfrac{6\sqrt{5}+4\sqrt{10}}{3}\)

d. \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\)

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

= \(\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)

= \(\left(-2\sqrt{3}-5+\dfrac{15+5\sqrt{3}}{2}\right).\dfrac{1}{\sqrt{3}+5}\)

= \(\dfrac{-4\sqrt{3}-10+15+5\sqrt{3}}{2}.\dfrac{1}{\sqrt{3}+5}\)

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

= \(\dfrac{1}{2}\)

Nếu đúng cho 1 like nhé!

30 tháng 9 2017

Rút gọn biểu thức chứa căn bậc hai

30 tháng 9 2017

Rút gọn biểu thức chứa căn bậc hai

13 tháng 6 2017

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)

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

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

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(5\sqrt{5}-11\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}\cdot\dfrac{1}{5\sqrt{5}-11}}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\left(\sqrt{5}-3\right)\cdot\left(5\sqrt{5}-1\right)}}\)

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

\(=\sqrt{\dfrac{2\sqrt{5}+6-5-3\sqrt{5}}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{\dfrac{-\sqrt{5}+1}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{-\dfrac{\left(-\sqrt{5}+1\right)\cdot\left(5\sqrt{5}+11\right)}{16}}\)

\(=\sqrt{-\dfrac{-25-11\sqrt{5}+5\sqrt{5}+11}{16}}\)

\(=\sqrt{-\dfrac{-14-6\sqrt{5}}{16}}\)

\(=\sqrt{-\dfrac{2\left(-7-3\sqrt{5}\right)}{16}}\)

\(=\sqrt{-\dfrac{-7-3\sqrt{5}}{8}}\)

\(=\dfrac{\sqrt{-\left(-7-3\sqrt{5}\right)}}{\sqrt{8}}\)

\(=\dfrac{\sqrt{7+3\sqrt{5}}}{2\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(7+3\sqrt{5}\right)\cdot2}}{4}\)

\(=\dfrac{\sqrt{14+6\sqrt{5}}}{4}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}}{4}\)

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

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

\(=\left(2+3\sqrt{5}\right)\cdot\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\cdot\left(\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(3-\sqrt{5}\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-3+\sqrt{5}\)

\(=9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

\(=\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

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

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

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

\(=\dfrac{2-1}{2}\)

\(=\dfrac{1}{2}\)

13 tháng 6 2017

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)= \(\dfrac{\sqrt{2-\sqrt{5}}}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}\sqrt{2-\sqrt{5}}}\)

= \(\dfrac{1}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}}\) = \(\dfrac{1}{\sqrt{\sqrt{5}-3}^2}\) = \(\dfrac{1}{3-\sqrt{5}}\)

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

= \(\dfrac{2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{8\sqrt{5}+19-5+2\sqrt{5}-\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\) = \(\dfrac{9\sqrt{5}+16}{5-4}\) = \(9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\) = \(\dfrac{1+\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

= \(\dfrac{1}{2}\)

2 tháng 9 2017

1/

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

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

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

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

\(=\sqrt{2}\)

2/

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

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

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

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

\(=-\left(5-2\right)=-3\)

#F.C

3 tháng 9 2017

máy câu còn lại thì sao

14 tháng 11 2018

Đề không khó, mỗi tội dài

14 tháng 11 2018

vậy thì bn làm hộ mik vs , mik cần gấp

25 tháng 7 2017

Hỏi đáp Toán

25 tháng 7 2017

Hỏi đáp Toán