`D=\frac{4^7.2^8.(-2)^5}{3.2^15.16^2-5.2^2.1024^2}`
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ta có: 4^7.2^8 / 3.2^15.16^2 - 5.2^2.(2^10)^2
= 2^14.2^8 / 3.2^15.2^8 - 5.2^2.2^20
= 2^22 / 6.2^22 - 5.2^22
= 2^22 / 2^22 ( 6 - 5 )
= 1
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(B=\frac{7}{2.7}\)
\(B=\frac{1}{2}\)
\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)
\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)
\(C=\frac{-2}{9}\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)
\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)
\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)
\(D=1\)
Ta có: \(\dfrac{4^7\cdot2^8}{3\cdot2^{15}\cdot16^2-5\cdot2^2\cdot\left(2^{10}\right)^2}\)
\(=\dfrac{2^{22}}{3\cdot2^{15}\cdot2^8-5\cdot2^2\cdot2^{20}}\)
\(=\dfrac{2^{22}}{2^{22}\left(3\cdot2-5\right)}\)
\(=1\)
C=\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{1}{2}=0,5\)
D=\(\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)\(=1\)
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