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http://olm.vn/hoi-dap/question/425436.html
\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}=\frac{5^2.6^{11}.6^2.6^2.6^6.2^6.3^2.5^2}{2.2^{12}.3^{12}2^4.5^4-3^8.2^{15}.3^3}=\frac{5^6.6^{21}.2^8.3^2}{2^{17}.3^{12}.5^4-3^{11}.2^{15}}=\frac{5^6.3^{23}.2^{29}}{3^{11}.2^{15}.\left(2^2.3.5^4-1\right)}=\frac{5^6.3^{13}.2^{14}}{2^2.3.5^4-1}\)
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Ta có:
\(\frac{5^2.6^{11}.16^6.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\Leftrightarrow\frac{5^2.6^{11}.16^6.12^2.15^2}{2.6^{11}.6.10^4-81^2.960^3}\)
\(\Leftrightarrow\frac{5^2.16^6.12^2.15}{2.6.10^4-81^2.960^3}\Leftrightarrow\frac{5^2.16^2.12.12.15}{\left(2.6\right).10^4-81^2.960^3}\)
\(\Leftrightarrow\frac{5^2.16^2.12.12.15}{12.10^4-81^2.960^3}\Leftrightarrow\frac{5^2.16^2.12.15}{10^4-81^2.960}\)
\(\Leftrightarrow\frac{5^2.16^2.12.3.5}{10^4-3^8.960}\Leftrightarrow\frac{5^2.16^2.12.3.5}{10^4-3^7.3.960}\Leftrightarrow\frac{5^2.16^2.12.5}{10^4-3^7.960}\)
Ps: Không chắc chắn đúng! Thầy cũng cho mình làm bài này hôm nay. Mình cũng làm cách tương tự như trên nhưng chưa biết đúng hay sai! Bạn thông cảm
làm cách trình bày ra nhé biết kết quả là \(\frac{388}{7199}\)
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\(=\frac{5.2^{30}.3^{18}-3^{20}.2^2.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}\cdot3^{18}}=\frac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2\left(10-9\right)}{\left(15-14\right)}==\frac{2.1}{1}=2\)
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Ta có: \(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
\(=\frac{5^2.2^{11}.3^{11}.2^8+2^2.3^2.3^6.2^{12}.5^2.3^5}{2.2^{12}.3^{12}.2^4.5^4-3^8.2^{18}.3^3.5^3}\)
\(=\frac{5^2.2^{19}.3^{11}+3^{13}.2^{14}.5^2}{2^{17}.3^{12}.5^4-2^{18}.3^{11}.5^3}\)
\(=\frac{\left(5^2.2^{14}.3^{11}\right)\left(2^5+3^2\right)}{\left(2^{17}.3^{11}.5^3\right)\left(3.5-2\right)}\)
\(=\frac{\left(2^5+3^2\right)}{\left(2^3.5\right)\left(3.5-2\right)}\)
\(=\frac{32+9}{40.13}\)
\(=\frac{41}{520}\)
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$\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2960^3}$52.611.162+62.126.1522.612.104−8129603= \(y=\frac{1}{x^2+\sqrt{x}}\)
\(y=\frac{1}{x^2+\sqrt{x}}\)
\(A=\dfrac{3^{32}\cdot2^{33}}{2^{33}\cdot3^{33}}=\dfrac{1}{3}\)
\(B=\dfrac{3^6\cdot5^{10}\cdot2^5}{2\cdot5^{14}\cdot3^{14}\cdot3^2\cdot2}=\dfrac{2^3}{3^8\cdot5^4}\)