có ai biết giải bài này k hộ mình vs ( chi tiết hộ mình nhé )
bài 1: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
b, \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
bài 2: trục căn thức ở mẫu và rút gọn
a, \(\dfrac{\sqrt{8}}{\sqrt{5}-\sqrt{3}}\)
b, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
bài 3: trục căn thức và thực hiện phép tính
a, M=\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
b, N= \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
Bài 3:
a.
\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)
\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)
\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)
b.
\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)
\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)
1a) \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}=\dfrac{2\sqrt{2}+3\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{2\sqrt{2}+3\sqrt{3}}{\left(2\sqrt{2}\right)^2-\left(3\sqrt{3}\right)^2}\)
\(=\dfrac{2\sqrt{2}+3\sqrt{3}}{-19}=-\dfrac{2\sqrt{2}+3\sqrt{3}}{19}\)
b) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{3-\sqrt{5}}{\sqrt{4}}=\dfrac{3-\sqrt{5}}{2}\)
2a) \(\dfrac{\sqrt{8}}{\sqrt{5}-\sqrt{3}}=\dfrac{2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\left(\sqrt{10}+\sqrt{6}\right)}{5-3}=\sqrt{10}+\sqrt{6}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{2-\sqrt{3}}{\sqrt{1}}=2-\sqrt{3}\)
Bài 1:
a) \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
\(=\dfrac{2\sqrt{2}+3\sqrt{3}}{8-27}=\dfrac{-2\sqrt{2}-3\sqrt{3}}{19}\)
b) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
Bài 3:
a) Ta có: \(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(11+\sqrt{6}\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(11+\sqrt{6}\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121
=-115
b) Ta có: \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(-\sqrt{5}+1\right)\left(-\sqrt{5}-1\right)\)
=5-1
=4
Bài 2:
a) Ta có: \(\dfrac{\sqrt{8}}{\sqrt{5}-\sqrt{3}}\)
\(=\dfrac{2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{2}\)
\(=\sqrt{10}+\sqrt{6}\)
b) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}\)
