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

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

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

a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)

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) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)

Các bạn vào trang cá nhân của mik đi, có cái này hay lắm!!!

\(a)\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)

\(b)\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)

\(c)\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^6\left(3^{25}+2^{30}\right)}{2^{11}.3^6\left(3^{25}+2^{30}\right)}=\frac{1}{2}\)

\(d)a\left(-b\right)\left(-a\right)^2\left(-b\right)^3\left(-a\right)^3\left(-b\right)^4\)

\(=\left[a\left(-a\right)^2\left(-a\right)^3\right]\left[\left(-b\right)\left(-b\right)^3\left(-b\right)^4\right]\)

\(=\left[a.a^2\left(-a\right)^3\right]\left[-b.b^4\left(-b\right)^3\right]=\left(-a^3.a^3\right)\left[\left(-b\right)^4.b^4\right]=-b^8\)

\(e)\left[-\left(-a\right)^3\right]\left(-a^2\right)^3\left[\left(-b\right)^2\right]^3\left[-\left(-b\right)^4\right]\)

\(=a^3\left(-a^6\right).b^6\left(-b^4\right)=-a^3.\left(-b^2\right)=a^3b^2\)

a,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)

\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)

\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)

\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)

\(=\frac{-8}{2005}\)

b,Ta có: \(81^{10}-27^{13}-9^{21}\)

\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)

\(=3^{40}-3^{39}-3^{42}\)

\(=3^{39}.3-3^{39}-3^{39}.3^3\)

\(=3^{39}.\left(3-1-3^3\right)\)

\(=3^2.3^{37}.\left(-25\right)\)

\(=3^{37}.\left(-225\right)⋮225\)

Vậy \(81^{10}-27^{13}-9^{21}⋮225\)

kệ cha nhà bây

a)   378

b)   3

c)    2

d)    2

e)    \(\frac{8748}{1715}\)

Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?