\(\sqrt{30}\)

Bằng 5,490838311

Đ/S: 5,490838311

\(\sqrt{\frac{9}{\sqrt{14+4\sqrt{6}}}}-\sqrt{\frac{9}{\sqrt{14-4\sqrt{6}}}}\)

\(=\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2+2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}-\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2-2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}\)

\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)

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

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

\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{12-2}}\)

\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{10}}\)

\(=\frac{3\left(\sqrt{2\sqrt{3}-\sqrt{2}}-\sqrt{2\sqrt{3}+\sqrt{2}}\right)}{\sqrt{10}}\)

\(=\frac{3}{\sqrt{10}}\)

\(\sqrt{\frac{9}{\sqrt{14+4\sqrt{6}}}}-\sqrt{\frac{9}{\sqrt{14-4\sqrt{6}}}}\)

\(=\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2+2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}-\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2-2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}\)

\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)

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

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

\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{12-2}}\)\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{10}}\)

\(=\frac{3\left(\sqrt{2\sqrt{3}-\sqrt{2}}-\sqrt{2\sqrt{3}+\sqrt{2}}\right)}{\sqrt{10}}\)

bí....!!!

\(\sqrt{\frac{9}{\sqrt{14+4\sqrt{6}}}}-\sqrt{\frac{9}{\sqrt{14-4\sqrt{6}}}}\)

\(=\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2+2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}-\sqrt{\frac{9}{\sqrt{\left(\sqrt{12}\right)^2-2\cdot\sqrt{12}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}}}\)

\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)

\(=\sqrt{\frac{9}{\sqrt{12}+\sqrt{2}}}-\sqrt{\frac{9}{\sqrt{12}-\sqrt{2}}}\)

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

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

\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{12-2}}\)

\(=\frac{3\sqrt{\sqrt{12}-\sqrt{2}}-3\sqrt{\sqrt{12}+\sqrt{2}}}{\sqrt{10}}\)

\(=\frac{3\left(\sqrt{2\sqrt{3}-\sqrt{2}}-\sqrt{2\sqrt{3}+\sqrt{2}}\right)}{\sqrt{10}}\)

\(=\frac{3.\left(-\sqrt{2\sqrt{3}-\sqrt{2}}\right)}{\sqrt{10}}\)

\(=\frac{3}{\sqrt{10}}\)

Lời giải:
\(P=\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)

\(=\sqrt{(7+2\sqrt{7.5}+5)+2(\sqrt{10}+\sqrt{14})+2}\)

\(=\sqrt{(\sqrt{7}+\sqrt{5})^2+2\sqrt{2}(\sqrt{5}+\sqrt{7})+(\sqrt{2})^2}\)

\(=\sqrt{(\sqrt{5}+\sqrt{7}+\sqrt{2})^2}=\sqrt{5}+\sqrt{7}+\sqrt{2}\)

\(\left(\frac{4\sqrt{a}}{\sqrt{a}+2}+\frac{8a}{4-a}\right):\left(\frac{\sqrt{a}-1}{a-2\sqrt{a}}-\frac{2}{\sqrt{a}}\right)\) (ĐKXĐ : \(a>0;a\ne4;a\ne9\))

\(=\left[\frac{4\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{8a}{a-4}\right]:\left[\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-2\right)}-\frac{2\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}\right]\)

\(=\frac{4a-8\sqrt{a}-8a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{\sqrt{a}-1-2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-2\right)}\)

\(=\frac{-4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-4\sqrt{a}}{\sqrt{a}-2}.\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{3-\sqrt{a}}=-\frac{4a}{3-\sqrt{a}}\)

\(\sqrt{3+\sqrt{5}}-\sqrt{4+\sqrt{7}}+\frac{\sqrt{14}}{2}\)

\(\approx1,581\)

=1,5811

Đặt  \(A=\sqrt{3+\sqrt{5}}\)\(-\sqrt{4+\sqrt{7}}\)\(+\frac{\sqrt{14}}{2}\)

\(A\sqrt{2}=\)\(\sqrt{6+2\sqrt{5}}\)\(-\sqrt{8+2\sqrt{7}}\)\(+\sqrt{14}\)

\(A\sqrt{2}=\)\(\sqrt{\left(\sqrt{5}+1\right)^2}\)\(-\sqrt{\left(\sqrt{7}+1\right)^2}\)\(+\sqrt{14}\)

\(A\sqrt{2}=\)\(\sqrt{5}+1-\sqrt{7}-1+\sqrt{14}\)

\(A\sqrt{2}=\)\(\sqrt{5}-\sqrt{7}+\sqrt{14}\)

\(A=\frac{\sqrt{5}}{\sqrt{2}}-\frac{\sqrt{7}}{\sqrt{2}}+\sqrt{7}\)

\(A=\frac{\sqrt{10}-\sqrt{14}}{2}+\sqrt{7}\)

A=\(\sqrt{\left(\sqrt{7}-2\right)^2}\)+\(\frac{25\sqrt{7}-63}{3\sqrt{7}-7}\)=\(\frac{12\sqrt{7}-28}{3\sqrt{7}-7}\)=4

a)    Ta có:

\(\begin{array}{l}10 + \left( { - 12} \right) =  - 2\\ - 2 + \left( { - 12} \right) =  - 14\\ - 14 + \left( { - 12} \right) =  - 26\\ - 26 + \left( { - 12} \right) =  - 38\end{array}\)

Dãy số là cấp số cộng

b)    Ta có:

\(\begin{array}{l}\frac{1}{2} + \frac{3}{4} = \frac{5}{4}\\\frac{5}{4} + \frac{3}{4} = 2\\2 + \frac{3}{4} = \frac{{11}}{4}\\\frac{{11}}{4} + \frac{3}{4} = \frac{7}{2}\end{array}\)

 Dãy số là cấp số cộng

c)    Không xác định được d giữa các số hạng

 Dãy số không là cấp số cộng

d)    Ta có:

 \(\begin{array}{l}1 + 3 = 4\\4 + 3 = 7\\7 + 3 = 10\\10 + 3 = 13\end{array}\)

Dãy số là cấp số cộng