Căn bậc hai. Căn bậc ba - Hỏi đáp

\(A=\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\dfrac{\sqrt{7-2\sqrt{6}}-1}{7-2\sqrt{6}-1}-\dfrac{\sqrt{7+2\sqrt{6}}-1}{7+2\sqrt{6}-1}\)

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

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

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

\(=\dfrac{2}{\sqrt{12}\left(\sqrt{3}+\sqrt{2}\right)}=\dfrac{2\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}\left(3-2\right)}=\dfrac{3-\sqrt{6}}{3}\)

\(5-2\sqrt{6}=\left(\sqrt{2}\right)^2-2\times\sqrt{2}\times\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

\(7+2\sqrt{10}=\left(\sqrt{2}\right)^2+2\times\sqrt{2}\times\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{2}+\sqrt{5}\right)^2\)

\(8-2\sqrt{15}=\left(\sqrt{5}\right)^3-2\times\sqrt{5}\times\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(B=\dfrac{2}{\sqrt{8-2\sqrt{15}}}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{3}{\sqrt{7+2\sqrt{10}}}\)

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

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

\(=\sqrt{5}+\sqrt{3}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}=0\)

a) Để \(M\in Z\)

thì \(\sqrt{a}+2⋮\sqrt{a}-2\)

\(\Rightarrow\left(\sqrt{a}-2\right)+4⋮\sqrt{a}-2\)

mà \(\sqrt{a}-2⋮\sqrt{a}-2\)

\(\Rightarrow4⋮\sqrt{a}-2\)

\(\Rightarrow\sqrt{a}-2\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{a}-2\in\left\{\pm1;\pm2;\pm4\right\}\)

......

b) Tìm ra các trường hợp a là số hữu tỉ ở câu a).

a) Ta có : ĐKXĐ :\(a\ne4;a\ge0\)

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

Mà \(A\in Z\) => \(1+\dfrac{4}{\sqrt{a}-2}\in Z=>\dfrac{4}{\sqrt{a}-2}\in Z=>\sqrt{a}-2\inƯ\left(4\right)\)

=>\(\left[{}\begin{matrix}\sqrt{a}-2=1\\\sqrt{a}-2=-1\\\sqrt{a}-2=4\\\sqrt{a}-2=-4\end{matrix}\right.< =>\left[{}\begin{matrix}a=9\\a=1\\a=36\\Loại\end{matrix}\right.\)

\(A=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{8}\right)2\sqrt{6}-5\sqrt{3}=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right)2\sqrt{6}-5\sqrt{3}=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right)2\sqrt{6}-5\sqrt{3}=12-18\sqrt{2}+16\sqrt{3}-5\sqrt{3}=12-18\sqrt{2}+11\sqrt{3}\)

\(B=\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}=12-12\sqrt{6}+18+2\sqrt{6}+6\sqrt{6}=30-4\sqrt{6}\)

a) Ta có :

4 - 2\(\sqrt{3}\) = 1 - 2.1.\(\sqrt{3}\) + 3 = 1 - 2.1.\(\sqrt{3}\) + (\(\sqrt{3}\))² = (1 - \(\sqrt{3}\))²= (\(\sqrt{3}\) - 1)²

b) Áp dụng câu a ta có:

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

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