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B= {[ 5 * (2^2)^15 * (3^2)^9 ] - [ (2^2) * 3^20 * (2^3)^9 ]} / {[ 5 * (2^9) * (2^19)*(3^19) ] - [ 7 * (2^29) * (3^3)^6 ]}
B= {[ 5 * (2^30) * (3^18) ] - [ (3^20) * (2^29) ]} / {[ 5 * (2^28) * (3^19) ] - [ 7 * (2^29) * (3^18) ]}
B= {[ (2^29) * (3^18) ] * [(5 * 2) - 3^2 ]} / {[ (2^28) * (3^18) ] - [(5 * 3) - (7 * 2)] }
B= [ (2^29) * (3^18) ] / [ (2^28) * (3^18) ]
B= [ (2^1) * (2^28) * (3^18) ] / [ (2^28) * (3^18) ]
B = 2
ta có \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)=\(\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)=\(\frac{2^{29}.3^{18}\left(2.5-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}\)=2
\(\left(-4,3.1,1+1,1.4,5\right):\left(-0,5:0,05+10,01\right)\)
\(=\left[1,1\left(-4,3+4,5\right)\right]:\left(-10+10,01\right)\)
\(=\left[1,1.0,2\right]:\left(0,01\right)\)
\(=0,22:0,01\)
\(=22\)
TA có\(\frac{2^{12}.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3}\)
=\(\frac{2^{12}.3^5.2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3}\)
=\(\frac{2^{24}.3^9}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3}\)
=\(2^{20}.3^4-30\)
a) \(\left(1,25\right)^3.8^3=\left(1,25.8\right)^3=1000\)
b) \(\left(\dfrac{-11}{9}\right)^4.\left(\dfrac{27}{22}\right)^4=\left(\dfrac{-11}{9}.\dfrac{27}{22}\right)^4=\left(\dfrac{-11.9.3}{9.2.\left(-11\right)}\right)^4\)
\(=\left(\dfrac{3}{2}\right)^4=\dfrac{81}{16}\)
c) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)
d) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=100^{-1}=0,01\)
\(=3\)