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

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

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

So so hang la 2011.

=> A = (-1)^2011=-1

kết quả bằng 200 nha

\(\frac{4^2\cdot4^3}{2^{10}}\)

C1: \(=\frac{\left(2^2\right)^2\cdot\left(2^2\right)^3}{2^{10}}\)

\(=\frac{2^4\cdot2^6}{2^{10}}\)

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

C2 : \(=\frac{4^{2+3}}{4^5}=\frac{4^5}{4^5}=1\)

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

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

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

A​=10^2-(5^2.4-4^3.3)+2^3=200 Tích cho mình nha

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 :)

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

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