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13 tháng 9 2015

Phạm Hồng Phúc lấy máy tính mà tính á bn

Bài 2: 

a: \(\left(0.25\right)^3\cdot32=\dfrac{1}{4^3}\cdot32=\dfrac{32}{64}=\dfrac{1}{2}\)

b: \(\left(-0.125\right)^3\cdot80^4=\left(-0.125\cdot80\right)^3\cdot80=-80\)

c: \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)

d: \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

7 tháng 6 2016

a) =\(\left(\frac{1}{4}\right)^3\cdot2^5=\frac{1}{2^6}2^5=0,5\)

b) \(=\left(\frac{1}{8}\right)^3\cdot8^4\cdot10^4=80.000\)

c) \(=8^2\cdot\frac{4^5}{2^{20}}=\frac{2^6\cdot2^{10}}{2^{20}}=\frac{1}{2^4}=0,0625\)

d) = \(\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3\)

22 tháng 9 2018

a) \(\left(0,25\right)^3.32=0,015625.32=0,5\)

b) \(\left(0,125\right)^3.80^4=0,001953125.40960000=80000\)

c) \(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}=\frac{2^5.2^{10}}{2^{20}}=\frac{2^{15}}{2^{20}}=\frac{1}{2^5}=\frac{1}{32}\)

d) \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{19}.\left(3^2\right)^{15}}=\frac{3^{44}.3^{17}}{3^{57}.3^{30}}=\frac{3^{61}}{3^{87}}=\frac{1}{3^{26}}\)

22 tháng 9 2018

siêu sao đá bóng sai c và d rồi

kết quả của c=\(\frac{1}{16}\)  kết quả của d=3

3 tháng 7 2016

a.

\(\left(0,25\right)^3\times32\)

\(=\left(0,25\right)^3\times2^5\)

\(=\left(0,25\right)^3\times2^3\times2^2\) 

\(=\left(0,25\times2\right)^3\times4\)

\(=\left(0,5\right)^3\times4\)

\(=0,125\times4\)

\(=0,5\)

b.

\(\left(-0,125\right)^3\times80^4\)

\(=\left(-0,125\right)^3\times80^3\times80\)

\(=\left(-0,125\times80\right)^3\times80\)

\(=\left(-10\right)^3\times80\)

\(=-1000\times80\)

\(=-80000\)

c.

\(3^{1994}+3^{1993}-3^{1992}\)

\(=3^{1992}\times\left(3^2+3-1\right)\)

\(=3^{1992}\times\left(9+3-1\right)\)

\(=3^{1992}\times11\)

\(\Rightarrow3^{1994}+3^{1993}-3^{1992}⋮11\)

d.

\(4^{13}+32^5-8^8\)

\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)

\(=2^{26}+2^{25}-2^{24}\)

\(=2^{24}\times\left(2^2+2-1\right)\)

\(=2^{24}\times\left(4+2-1\right)\)

\(=2^{24}\times5\)

\(\Rightarrow4^{13}+32^5-8^8⋮5\)

Chúc bạn học tốtok

4 tháng 7 2016

thanhks bạn nhiều 

19 tháng 10 2021

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

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

a) \(\left(0,25\right)^3\cdot32=0,015625\cdot32=0,5\)

b) \(\left(-0,125\right)^3\cdot80^4=\dfrac{-1}{512}\cdot40960000=80000\)

c) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^{3^2}\cdot2^{2^5}}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)

d) \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{4^{11}}\cdot3^{17}}{3^{3^{10}}\cdot3^{2^{15}}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)