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2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
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!
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) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{1+\sqrt{2}}{\sqrt{3}-1}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\left(1+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\left(\sqrt{2}+1\right)\cdot\left(\sqrt{2}-1\right)}{3-1}\)
\(=\dfrac{2-1}{2}\)
\(=\dfrac{1}{2}\)
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) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\) = \(\dfrac{1+\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
= \(\dfrac{1+\sqrt{2}}{\sqrt{3}-1}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\) = \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\) = \(\dfrac{\sqrt{2}-1+2-\sqrt{2}}{3-1}\)
= \(\dfrac{1}{2}\)
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 ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$