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a) \(A=\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(A=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(A=\dfrac{11.3^{29}-3^{29}.3}{2^2.3^{28}}\)
\(A=\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)
\(A=\dfrac{3^{29}.8}{2^2.3^{28}}\)
\(A=\dfrac{3.8}{4}=6\)
vậy \(A=6\)
b) \(B=\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(B=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(B=\dfrac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(B=\dfrac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)
\(B=\dfrac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(B=\dfrac{3^2.2^{36}}{2^{35}.9}\)
\(B=\dfrac{3^2.2}{9}\)
\(B=\dfrac{9.2}{9}\)
\(B=2\)
vậy \(B=2\)
thank bạn
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)