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\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}}{3}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2-\sqrt{5}-2\)
\(=-\sqrt{5}\)
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.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{2}\)
4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(=-3-3\sqrt{3}-3\)
\(=-6-3\sqrt{3}\)
a) \(=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(=\dfrac{3\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=3\sqrt{2}+3+\sqrt{2}=3+4\sqrt{2}\)
c) \(=\dfrac{3\left(\sqrt{5}+2\right)-3\left(\sqrt{5}-2\right)}{5-4}=3\sqrt{5}+6-3\sqrt{5}+6=12\)
\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}\right)^2-\left(\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)^2\)
\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}+\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left[\dfrac{2\left(2-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{2\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}\right)+2\left(2\sqrt{3}-2+3-\sqrt{3}\right)}{3-1}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}+2\sqrt{3}+1-\sqrt{3}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}-2\sqrt{3}+2-3+\sqrt{3}\right)}{2}\right]\)
\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1-\sqrt{15}-\sqrt{5}\right)\)
\(A=\left(\sqrt{3}+1\right)^2\left(3-\sqrt{5}\right)\left(1-\sqrt{5}\right)\)
bạn cho mình hỏi 2 dòng cuối sao lại ra vậy thế