\(\frac{11.3^{22}.3^7-19^{15}}{\left(2.3^{14}\right)^2}\)
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\(=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{14.2}}\)
\(=\frac{11.3^{29}-3.3^{29}}{2^2.3^{28}}\)
\(=\frac{\left(11-3\right).3^{29}}{2^2.3^{28}}\)
\(=\frac{8.3^{29}}{4.3^{28}}\)
\(=6\)
ta có \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}\)
= \(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
= \(\frac{3^{29}.2^3}{2^2.3^{28}}=3.2=6\)
\(\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{2^2.3^{28}}=\frac{3^{28}.3.2^3}{2^2.3^{28}}=6\)
\(=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}=\frac{11.3^{29}-3^{30}}{4.3^{28}}=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}=\frac{3.8}{4}=6\)