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a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có a2 + b2 = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a2 - b2 = \(2\sqrt{5}\)
Ta có cái ban đầu
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=
\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)
a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\left(\sqrt{10}+\sqrt{2}\right)\)
\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)
\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)
\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)
b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right)\)
\(=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5+2\sqrt{5}+1}.\left(\sqrt{5}-1\right)\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)
\(=2.\left(5-1\right)=2.4=8\)
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)
d) \(\frac{1}{\sqrt{3}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}-\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{2\sqrt{5}}{3-5}=\frac{2\sqrt{5}}{-2}=-\sqrt{5}\)c) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}+\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
b) \(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}+\sqrt{5-2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)a) \(\sqrt{27}+\sqrt{243}-6\sqrt{12}=\sqrt{9.3}+\sqrt{81.3}-6\sqrt{4.3}=3\sqrt{3}+9\sqrt{3}-12\sqrt{3}=0\)
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2}{2}\)
c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))
\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-3}{3}\)
b) Tử \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\) (hằng đăngt thức số 3 )
\(\text{Theo đề bài: }=\dfrac{3\sqrt{2}+6\sqrt{3}+2\sqrt{5}-\sqrt{6}}{2}\)