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\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.30^{14}\right)^2}=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.30^{28}}=\frac{11.3^{29}-3^{30}}{2^2.\left(3.10\right)^{28}}=\frac{11.3^{29}-3.3^{29}}{2^2.10^{28}.3^{28}}\)
\(=\frac{3^{29}\left(11-3\right)}{2^2.10^{28}.3^{28}}=\frac{3.2}{10^{28}}=\frac{6}{10^{28}}\)
\(\frac{11\times3^{22}\times3^7\times9^{15}}{\left(2\times3^{14}\right)^2}=\frac{11\times3^{29}\times\left(3^2\right)^{15}}{2^2\times3^{28}}\)\(=\frac{11\times3^{29}\times3^{30}}{4\times3^{28}}=\frac{11\times3^{59}}{4\times3^{28}}\)\(=\frac{11\times3^{31}}{4}\)
- Kết quả to quá nên bạn tưn tính nhé !
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
a)\(3^9:3^7+5\times2^3\)
\(9+40=49\)
b)\(2^3\times3^2-5^{16}:5^{14}\)
\(72-25=47\)
Ta có : \(A=11.3^{22}.3^7-9^{15}\)
\(=11.3^{29}-\left(3^2\right)^{15}\)
\(=11.3^{29}-3^{30}\)
\(=3^{29}\left(11-3\right)\)
\(=3^{29}.8\)
Ta có : \(B=\left(2.3^{14}\right)^2=2^2.3^{28}\)
câu a và b là 1 mà
cách làm đúng nhưng ko phải tách ra đâu nha
