a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)

\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)

\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)

b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)

\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)

c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)

\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)

e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)

\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)

\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)

\(B=1-\left(\dfrac{1}{2}\right)^{99}\)

\(2,\)

\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)

\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)

\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)

\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)

\(=\dfrac{3^5.2^{10}}{5^{20}}\)

\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)

\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)

\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)

\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)

\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)

\(3,\)

\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)

\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

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, \(5^{x+2}=625\)

\(5^{x+2}=5^4\)

\(x+2=4\)

\(x=2\)

a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)

8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)

=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)

=7.(-7)

=-49

a: \(\dfrac{5^5}{5^x}=5^{18}\)

=>5-x=18

hay x=-13

b: \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Leftrightarrow2^{4-x}=\left(2^5\right)^6\cdot\left(2^4\right)^5=2^{30+20}=2^{50}\)

=>4-x=50

hay x=-46

c: \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Leftrightarrow2^{2x-3}=2^9\cdot2^{20}\cdot2^{20}=2^{49}\)

=>2x-3=49

=>2x=52

hay x=26

d: \(\dfrac{2^3}{2^x}=4^5\)

\(\Leftrightarrow2^{3-x}=2^{10}\)

=>3-x=10

hay x=-7

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)

\(\Leftrightarrow5^x=5^6\)

hay x=6

f: \(7\cdot2^x=2^9+5\cdot2^8\)

\(\Leftrightarrow2^x\cdot7=2^8\cdot7\)

\(\Leftrightarrow2^x=2^8\)

hay x=8

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) \(\dfrac{ \left(x^2\right)^5}{\left(x^5\right)^2}=\dfrac{x^{10}}{x^{10}}=1\)

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!!!

\(\text{a) }\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(5\cdot4\right)^4}{\left(5^2\right)^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{5^8\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{25\cdot4}=\dfrac{1}{100}\)

\(\text{b) }\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{2^7\cdot3^6}{2^{11}\cdot3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

\(\text{c) }\dfrac{45^{10}\cdot5^{20}}{75^5}=\dfrac{\left(5\cdot9\right)^{10}\cdot5^{20}}{\left(25\cdot3\right)^5}=\dfrac{5^{10}\cdot9^{10}\cdot5^{20}}{25^5\cdot3^5}=\dfrac{5^{10}\cdot5^{20}\cdot\left(3^2\right)^{10}}{\left(5^2\right)^5\cdot3^5}=\dfrac{5^{30}\cdot3^{20}}{5^{10}\cdot3^5}=5^{20}\cdot3^{15}\)

\(\text{d) }\left(0.8\right)^5=\left(\dfrac{8}{10}\right)^5=\left(\dfrac{4}{5}\right)^5=\dfrac{4^5}{5^5}=\dfrac{64}{3125}\)

\(\text{e) }\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9\cdot3^6}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)

\(f\text{) }\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)

\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)

a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15

=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)

=1 - 3/4

=1/4

b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2

=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)

=-8 +3/2 +1/3 -3/10

=-97/15

cứ phan tích cho hết đi là đc 9^6. 9^10 = (3^2)^6...................

tự làm đi

1. Tính:

a. \(\dfrac{9^6.9^{10}}{3^{32}}=\dfrac{\left(3^2\right)^6.\left(3^2\right)^{10}}{3^{32}}=\dfrac{3^{12}.3^{20}}{3^{32}}=\dfrac{3^{32}}{3^{32}}=1\)

b. \(\dfrac{25^8.25^{10}}{5^{34}}=\dfrac{\left(5^2\right)^8.\left(5^2\right)^{10}}{5^{34}}=\dfrac{5^{16}.5^{20}}{5^{34}}=\dfrac{5^{36}}{5^{34}}=5^{36}:5^{34}=5^2=25\)

c. \(\dfrac{7^{56}}{49^9.49^{20}}=\dfrac{7^{56}}{\left(7^2\right)^9.\left(7^2\right)^{20}}=\dfrac{7^{56}}{7^{18}.7^{40}}=\dfrac{7^{56}}{7^{58}}=7^{56}:7^{58}=\dfrac{7^{56}}{7^{56}.7^2}=\dfrac{1}{7^2}=\dfrac{1}{49}\)

d. \(\dfrac{4^2.4^3}{2^{10}}=\dfrac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\dfrac{2^4.3^6}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)

e. \(\dfrac{2^{17}.25^5}{10^8.8^3}=\dfrac{2^{17}.\left(5^2\right)^5}{\left(2.5\right)^8.\left(2^3\right)^3}=\dfrac{2^{17}.5^{10}}{2^8.5^8.2^9}=\dfrac{2^{17}.5^{10}}{2^{17}.5^8}=\dfrac{5^{10}}{5^8}=5^{10}:5^8=5^2=25\)

f. \(\dfrac{3^{15}.25^4}{15^6.27^3}=\dfrac{3^{15}.\left(5^2\right)^4}{\left(3.5\right)^6.\left(3^3\right)^3}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{3^{15}.5^8}{5^6.3^6.3^9}=\dfrac{5^8}{5^6}=5^8:5^6=5^2=25\)

2. Tính lũy thừa âm:

a. 3^-2 = \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)

b. 2^-3 = \(\dfrac{1}{2^3}\) = \(\dfrac{1}{8}\)

3. Tính :

a. \(\dfrac{\left(0,8\right)^4}{\left(0,4\right)^3}=\dfrac{\left(0,8\right)^3.0,8}{\left(0,4\right)^3}=\left(\dfrac{0,8}{0,4}\right)^3.0,8=2^3.0,8=8.0,8=6,4\)

b. \(\dfrac{\left(0,8\right)^3}{\left(0,4\right)^4}=\dfrac{\left(0,8\right)^3}{\left(0,4\right)^3:0,4}=\left(\dfrac{0,8}{0,4}\right)^3.\dfrac{1}{0,4}=2^3.2,5=8.2,5=20\)

c. \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,6\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\left(\dfrac{\left(0,6\right)}{\left(0,2\right)}\right)^5.\dfrac{1}{0,2}=3^5.\dfrac{1}{0,2}=\dfrac{3^5}{0,2}=1215\)

P/s : Chế Mai Ngọc Trâm thử tham khảo thử đi!!!