*Tính
\(\frac{6^3+2\times6^2+2^3}{74}\)
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\(\frac{1.2.6.4.6.4.5.2}{2.3.6.8.6.2.2.2.8.10}=\frac{1}{96}\)
63+3x62+33/13
=63+31x62+33/13
=62x31x(6+32)/13
=1620/13
Em nghĩ là thế !
= (2.3)^3+ 3. ( 2.3)^2 + 3^3 / -13
= 2^3 .3^3 + 3. 3^2 . 2^2 + 3^3 / -13
= 3^3. ( 2^3 + 2^2 + 1) /-13
= 27.13/-13
= -27
\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^2\cdot3}{-13}=\frac{36\cdot9+9\cdot3}{-13}=\frac{39\cdot9}{-13}=-3\cdot9=-27\)
\(P=\dfrac{3^{140}\cdot2^{74}-2^4}{3^{72}\cdot3^{70}\cdot2^{70}-3^2}=\dfrac{2^4\left(3^{140}\cdot2^{70}-1\right)}{3^2\left(3^{140}\cdot2^{70}-1\right)}=\dfrac{16}{9}\)
\(\frac{5^2\times6^{11}\times16^2+6^2\times12^6\times15^2}{2\times6^{12}\times10^4-81^2\times960^3}\)
\(=\frac{5^2\times\left(2\times3\right)^{11}\times\left(2^4\right)^2+\left(2\times3\right)^2\times\left(2^2\times3\right)^6\times\left(3\times5\right)^2}{2\times\left(2\times3\right)^{12}\times\left(2\times5\right)^4-\left(3^4\right)^2\times\left(2^6\times3\times5\right)^3}\)
\(=\frac{5^2\times2^{19}\times3^{11}+2^{14}\times3^{10}\times5^3}{2^{17}\times5^4\times3^{12}-3^{11}\times2^{18}\times5^3}\)
\(=\frac{5^2\times3^{10}\times2^{14}\times\left(2^5\times3+5\right)}{2^{17}\times5^3\times3^{11}\times\left(5\times3-2\right)}\)
\(=\frac{2^5\times3+5}{2^3\times5\times3\times12}\)
\(=\frac{32\times3+5}{8\times15\times12}=\frac{96+5}{120\times12}=\frac{101}{1440}\)
\(\frac{\left(-0,25\right)^{-5}.9^4.\left(-2\right)^{-3}-2^{-2}.6^9}{2^9.3^6+6^6.40}\)
\(=\frac{\left(-4\right)^5.\left(3^2\right)^4.\left(-2\right)^{-3}-2^{-2}.\left(3.2\right)^9}{2^9.3^6+\left(2.3\right)^6.2^3.5}\)
\(=\frac{-\left(2^2\right)^5.3^8.\left(-2\right)^{-3}-2^{-2}.3^9.2^9}{2^9.3^6+2^6.3^6.2^3.5}\)
\(=\frac{-2^{10}3^8.\left(-2\right)^{-3}-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)
\(=\frac{2^73^8.-2^7.3^9}{2^9.3^6+2^9.3^6.5}\)
\(=\frac{2^7.3^8.\left(1-3\right)}{2^9.3^6.\left(1+5\right)}\)
\(=\frac{3^2.\left(-2\right)}{2^2.6}\)
\(=\frac{-3}{4}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)