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\(\frac{\left(4.3^{22}+7.3^{21}\right).57}{\left(19.27^4\right)^2}=\frac{3^{21}\left(4.3+7\right).57}{19^2.\left[\left(3^3\right)^4\right]^2}=\frac{3^{21}.19.57}{19^2.3^{24}}=\frac{3^{22}.19^2}{19^2.3^{24}}=\frac{1}{3^2}=\frac{1}{9}\)
\(=\frac{\left(0,5\right)^5.2^9}{2^6.2^4}=\frac{\left(0,5\right)^5.2^9}{2^9.2}=\left(\frac{1}{2}\right)^5\div2\)
\(=\frac{1^5}{2^5}.\frac{1}{2}=\frac{1}{2^6}=\frac{1}{64}\)
\(A=1+2+2^2+2^3+...+2^{10}\)
\(2A=2+2^2+2^3+2^4+...+2^{11}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)
\(A=2^{11}-1< 2^{11}\)
\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)
\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)
\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)
\(B=-2^{11}+2^3+10.2^{11}-2^3\)
\(B=9.2^{11}\)
Ta cần so sánh: \(9.2^{11}\) và \(2^{14}\)
Hay \(9\) và \(2^3\)
\(9>8=2^3\Leftrightarrow B>2^{14}\)
\(12^n:2^{2n}=3^n.\left(2^2\right)^n:2^{2n}=3^n.2^{2n}:2^{2n}=3^n\)
\(3^8:3^4+2^2.2^3=3^4+2^5=81+32=113\)
\(\left(7^{1997}-7^{1995}\right)\left(7^{1994}\cdot7\right)=7^{1995}\left(7^2-1\right)\cdot7^{1995}=7^{1995\cdot2}\cdot48=7^{3990}\cdot48\)
\(4^{14}\cdot5^{28}=4^{14}\cdot\left(5^2\right)^{14}=\left(4\cdot25\right)^{14}=100^{14}\)
\(3\cdot4^2-2\cdot3^2=3\cdot2^4-2\cdot3^2=6\left(2^3-3\right)=6\cdot5=30\)
\(18^3:9^3=\left(18:9\right)^3=2^3=8\)
\(\left(2^8+8^3\right):\left(2^5\cdot2^3\right)=\left(2^8+2^9\right):2^8=\dfrac{2^8}{2^8}+\dfrac{2^9}{2^8}=1+2=3\)
\(16\cdot64\cdot8^2:\left(4^3.2^5.16\right)=2^4\cdot2^6\cdot2^6:\left(2^6\cdot2^5\cdot2^4\right)=2\)
\(5\cdot2^9\cdot6^{10}-7\cdot2^{29}\cdot27^6=5\cdot2^9\cdot2^{10}\cdot3^{10}-7\cdot2^{29}\cdot3^{18}=2^{19}\cdot3^{10}\left(5\cdot3^{10}-7\cdot2^{10}\cdot8\right)\)
ai giúp mik với
là giá trị có tần số lớn nhất