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\(D=\left|\sqrt{8}-3\right|+\left|\sqrt{19}-4\right|-\left(\sqrt{19}-\sqrt{8}\right)\)
\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-\sqrt{8}\right)\)
\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)
\(D=-2\sqrt{2}+3+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)
\(D=-2\sqrt{2}+3+\sqrt{19}-4-\sqrt{19}+2\sqrt{2}\)
\(D=-2\sqrt{2}+3-4+2\sqrt{2}\)
\(D=3-4\)
\(D=-1\)
\(A=\frac{\sqrt{18}}{\sqrt{8}}+\frac{\sqrt{8}}{\sqrt{50}}=\frac{\sqrt{18}.\sqrt{50}+\sqrt{8}.\sqrt{8}}{\sqrt{8}.\sqrt{50}}=\frac{\sqrt{900}+\sqrt{64}}{\sqrt{400}}=\frac{30+8}{20}=\frac{38}{20}=\frac{19}{10}\)
\(A=\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\frac{2^{30}+2^{20}}{2^{22}+2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}\)\(=2^4=16.\)
#)Giải :
\(A=\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}}=\sqrt{\frac{2^{30}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}}=\sqrt{\frac{2^{30}}{2^{12}}}=\sqrt{2^8}=\sqrt{256}=16\)
\(A=\sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}=\sqrt{\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}}=\sqrt{2^{20-12}}=\sqrt{2^8}=2^4=16\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)
=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)
=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)
=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)
=\(-\dfrac{2}{3}\)
Ta có :
\(\sqrt{3}< \sqrt{4}=2\)
\(\sqrt{8}< \sqrt{9}=3\)
\(\sqrt{24}< \sqrt{25}=5\)
\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)
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