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\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
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: \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
a/ \(\left(-0,125\right)^3:80^4=-\frac{1}{512}.80^4=-80000\)
b/ \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=3\)
a, ( - 0, 125)^3 . 80^ 4
\(\left(-\frac{1}{8}\right)^3.80^4=\frac{-1^3}{8^3}\cdot8^4\cdot10^4\) = \(-8\cdot10^4=-8.1000=-8000\)
b,\(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{3^{4.11}.3^{17}}{3^{3.10}.3^{2.15}}=\frac{3^{44}.3^{17}}{3^{30}.3^{30}}=\frac{3^{44+17}}{3^{30+30}}=\frac{3^{61}}{3^{60}}=3\)
a,32
b,\(-\frac{1}{10}\)
c,-1000000
d,\(\frac{9}{16}\)
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)
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) \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{19}.\left(3^2\right)^{15}}=\frac{3^{44}.3^{17}}{3^{57}.3^{30}}=\frac{3^{61}}{3^{87}}=\frac{1}{3^{26}}\)
siêu sao đá bóng sai c và d rồi
kết quả của c=\(\frac{1}{16}\) kết quả của d=3