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a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)
b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)
c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)
d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)
e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)
\(A=\dfrac{4\cdot5^{10}\cdot5^{10}}{75^{10}}=\dfrac{4\cdot5^{20}}{\left(3\cdot25\right)^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot25^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot\left(5^2\right)^{10}}=\dfrac{4\cdot5^{20}}{3^{10}\cdot5^{20}}=\dfrac{4}{3^{10}}\)
\(B=\dfrac{\left(0.8\right)^5}{\left(0.4\right)^6}=\dfrac{\left(\dfrac{4}{5}\right)^5}{\left(\dfrac{2}{5}\right)^6}=\dfrac{\left(2\cdot\dfrac{2}{5}\right)^5}{\left(\dfrac{2}{5}\right)^6}=\dfrac{2^5\cdot\left(\dfrac{2}{5}\right)^5}{\left(\dfrac{2}{5}\right)^5\cdot\dfrac{2}{5}}=\dfrac{2^5}{\dfrac{2}{5}}=2^5\cdot\dfrac{5}{2}=\dfrac{32\cdot5}{2}=80\)
\(C=\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\dfrac{3^8}{3^6}=3^2=9\)
\(D=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{^{20}}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=226\)
Câu 1:\(\frac{45^{10}.5^{10}}{75^{10}}\) = \(\frac{\left(5.9\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\frac{5^{10}.9^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(\frac{9^{10}}{3^{10}}\) = \(\frac{3^{10}.3^{10}}{3^{10}}\) = \(3^{10}\) = 59049
Câu 2:\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4.2\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\) = \(\frac{2^5}{0,4}\) = \(\frac{32}{0,4}\) = 80
Câu 3:\(\frac{2^{15}.9^4}{6^3.8^3}\) = \(\frac{2^{15}.3^8}{2^{12}.3^3}\) = \(\frac{2^3.3^5}{1.1}\) = \(\frac{8.243}{1}\) = 1944
Câu 4: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\) = \(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\) = \(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\) = \(\frac{2^{20}.2^{10}+2^{20}}{2^{12}+2^{12}.2^{10}}\) = \(\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\) = \(\frac{2^{20}}{2^{12}}\) = \(\frac{2^8}{1}\) = \(2^8\) = 256
a) \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{3^6.2^6.2^9}=\dfrac{2^{15}.3^8}{3^6.2^{15}}=3^2=9\)
b) \(\dfrac{45^{15}.5^{15}}{75^{15}}=\dfrac{\left(9.5\right)^{15}.5^{15}}{\left(3.25\right)^{15}}=\dfrac{9^{15}.5^{15}.5^{15}}{3^{15}.25^{15}}=\dfrac{\left(3^2\right)^{15}.5^{30}}{3^{15}.\left(5^2\right)^{15}}\)
\(\dfrac{3^{30}.5^{30}}{3^{15}.5^{30}}=3^{15}=14348907\)
c) \(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}\)
\(=\dfrac{2^{20}}{2^{12}}=2^8=256\)
d) \(\dfrac{ \left(x^2\right)^5}{\left(x^5\right)^2}=\dfrac{x^{10}}{x^{10}}=1\)
((3\(^2\)))\(^2\) - ((-5\(^2\)))\(^2\) + ((-2\(^3\)))\(^2\)
= 81 - 625 + 64
= -544+ 64
= -480
2\(^4\) + 8[(-2)\(^2\) :\(\dfrac{1}{2}\)]\(^0\) - 2\(^{-2}\). 4 + (-2)\(^2\)
= 16+ 8.1 - \(\dfrac{1}{4}\). 4 + 4
= 16+ 8- 1+4
= 27
2\(^4\) + 3(\(\dfrac{1}{2}\))\(^0\) + 2\(^{-2}\).8 + [(-2)\(^3\). \(\dfrac{1}{2^4}\)].2 - \(\dfrac{1}{2}\)
= 16 + 3.1 +\(\dfrac{1}{4}\).8 + [(-8).\(\dfrac{1}{16}\)].2 -\(\dfrac{1}{2}\)
= 16 + 3+ 2 + \(\dfrac{-1}{2}\).2- \(\dfrac{1}{2}\)
= 21 + (-1)- \(\dfrac{1}{2}\)
= 20-\(\dfrac{1}{2}\) = \(\dfrac{40}{2}\) - \(\dfrac{1}{2}\)= \(\dfrac{39}{2}\)
\(\dfrac{15^{10}.5^{10}}{75^{10}}\) + \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
= \(\dfrac{\left(15.5\right)^{10}}{75^{10}}\) + \(\dfrac{\left(0,4.2\right)^5}{\left(0.4\right)^6}\)
= \(\dfrac{75^{10}}{75^{10}}\) + \(\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)
= 1 + \(\dfrac{2^5}{0,4}\) = 1+ 80 = 81
\(\dfrac{2^{13}.9^4}{6^3.8^3}\)
= \(\dfrac{2^{13}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}\) = \(\dfrac{2^{13}.3^8}{2^3.3^3.2^9}\)
= \(\dfrac{2^4.3^5}{2^3}\) = 2.3\(^5\) = 486
a, \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.2^8.5^4}{5^{10}.2^{10}}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)
b, \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
c, \(\dfrac{45^{10}.5^{20}}{75^5}=\dfrac{5^{10}.3^{20}.5^{20}}{3^5.5^{10}}=5^{20}.3^{15}\)
d, \(\left(0,8\right)^5=\left(0,1\right)^5.8^5=\dfrac{1}{100000}.32768=0,32768\)
e, \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{2^{15}.3^8}{2^6.3^6.2^9}=3^2=9\)
d, \(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=2^{10}=1024\)
Chúc bạn học tốt!!!
\(\text{a) }\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(5\cdot4\right)^4}{\left(5^2\right)^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{5^8\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{25\cdot4}=\dfrac{1}{100}\)
\(\text{b) }\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{2^7\cdot3^6}{2^{11}\cdot3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
\(\text{c) }\dfrac{45^{10}\cdot5^{20}}{75^5}=\dfrac{\left(5\cdot9\right)^{10}\cdot5^{20}}{\left(25\cdot3\right)^5}=\dfrac{5^{10}\cdot9^{10}\cdot5^{20}}{25^5\cdot3^5}=\dfrac{5^{10}\cdot5^{20}\cdot\left(3^2\right)^{10}}{\left(5^2\right)^5\cdot3^5}=\dfrac{5^{30}\cdot3^{20}}{5^{10}\cdot3^5}=5^{20}\cdot3^{15}\)
\(\text{d) }\left(0.8\right)^5=\left(\dfrac{8}{10}\right)^5=\left(\dfrac{4}{5}\right)^5=\dfrac{4^5}{5^5}=\dfrac{64}{3125}\)
\(\text{e) }\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9\cdot3^6}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)
\(f\text{) }\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)