\(\frac{4^2.4^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

\(\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{\left(2^2\right)^5}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}\)

               \(=\frac{2^4.2^6}{2^{10}}\)

                 \(=\frac{2^{10}}{2^{10}}=1\)

4^2.4^3/2^10=(2^2)^2/(2^2)^10=2^4/2^20=1/2^16

Nho cho minh nhe

Yen tam minh hoc roi khong sai dau

uk cảm ơn

hai vế mỗi vế có kết qua bằng 1

khi cộng 2 vầ ta có kết quả chính bằng 2

vậy thôi

dễ

\(\frac{4^2.4^3}{2^{10}}+\frac{3^2.3^3}{3^5}\)

\(=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}+\frac{3^{2+3}}{3^5}\)

\(=\frac{2^4.2^6}{2^{10}}+\frac{3^5}{3^5}\)

\(=\frac{2^{4+6}}{2^{10}}+1\)

\(=\frac{2^{10}}{2^{10}}+1\)

\(=1+1\)

\(=2\)

\(\frac{4^2.4^3}{2^{10}}+\frac{3^2.3^3}{3^5}=\frac{4^5}{2^{10}}+\frac{3^5}{3^5}=\frac{\left(2^2\right)^5}{2^{10}}+1=\frac{2^{10}}{2^{10}}+1=1+1=2\)

Chúc học tốt :)

a) \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

b) \(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{3^5.0,2^5}{0,2^6}=\frac{3^5}{0,2}=\frac{243}{0,2}=1215\)

c) \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3}{2^{11}}=\frac{3}{2^4}=\frac{3}{16}\)

d) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^2}{-13}=\frac{3^2\left(6^2+1\right)}{-13}=\frac{9.37}{-13}=\frac{333}{-13}\)

a,\(\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{4^5}{4^5}=1\)

\(\frac{\left(0.6\right)^5}{\left(0.2\right)^6}=1215\)

còn lại làm đc mà

\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

a, \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{4+6}}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

b,\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}\)

c, \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{3^5.2^{11}}=\frac{3}{2^4}\)

d, \(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(2.3\right)^3+3\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)

\(=\frac{2^3.3^3+3^3.2^2+3^3}{-13}=\frac{3^9\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3=27\)

a) \(\frac{4^2.4^3}{2^{10}}=\frac{4^{2+3}}{2^{10}}=\frac{\left(2^2\right)^5}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)    b) \(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(3.0,2\right)^5}{\left(0,2\right)^6}=\frac{3^5.0,2^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=1215\)

c) \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)               d)\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(2.3\right)^3+3.\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^2}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=-3^3\)

\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)

\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)