a: \(=\left(\dfrac{2}{18}-\dfrac{15}{18}-\dfrac{72}{18}\right):\left(\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}\right)\)

\(=\dfrac{-85}{18}:\dfrac{-170}{18}\)

\(=\dfrac{85}{170}=\dfrac{1}{2}\)

b: \(=\left(\dfrac{5}{8}-\dfrac{5}{6}-\dfrac{5}{32}+\dfrac{5}{64}\right):\left(1-\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}\right)\)

\(=\dfrac{-55}{192}:\dfrac{3}{8}=\dfrac{-55}{192}\cdot\dfrac{8}{3}=-\dfrac{55}{72}\)

\(\frac{11.3^{22}.3^7-9^{15}}{2^2.3^{28}}=\frac{11.3^{29}}{2^2.3^{28}}-\frac{3^{30}}{2^2.3^{28}}=\frac{11.3}{4}-\frac{9}{4}=\frac{33-9}{4}=\frac{24}{4}=6\)

\(\frac{27^5.8^2-9^7.4^3}{2^6.9^6}=\frac{3^{15}.2^6}{2^6.3^{12}}-\frac{3^{14}.2^6}{2^6.3^{12}}=27-9=18\)

~ Học tốt ~

a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)

• Đổi về cơ số 3: \(27 = 3^{3} , \textrm{ }\textrm{ } 81 = 3^{4}\).

\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)

b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)

• Đổi về cơ số 10: \(100 = 10^{2} , \textrm{ }\textrm{ } 1000 = 10^{3} , \textrm{ }\textrm{ } 10 \textrm{ } 000 = 10^{4}\).

\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)

c) \(\frac{36^{5}}{18^{5}}\)

• Do cùng số mũ:

\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)

d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)

\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)

e) \(\frac{125^{4}}{5^{8}}\)

• Đổi về cơ số 5: \(125 = 5^{3}\).

\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)

TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)

bạn muốn chép đáp án hay sem cách làm///??

a: \(3^7\cdot27^5\cdot81^3\)

\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)

\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)

b: \(100^6\cdot1000^5\cdot10000^3\)

\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)

\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)

c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)

d: \(24\cdot5^2+5^2\cdot5^3\)

\(=5^2\left(24+5^3\right)\)

\(=25\cdot\left(24+125\right)=25\cdot149=3725\)

e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)

Đăng lần lượt nhiều nhất 3 câu hỏi thôi, chứ đăng nhiều vậy, thấy nhát làm lắm. 

\(a,\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0.5-\frac{36}{21}\)

\(=\frac{11}{24}+\frac{13}{24}-\frac{5}{41}+\frac{1}{2}-\frac{36}{21}\)

\(=\frac{24}{24}-\frac{5}{41}+\frac{1}{2}-\frac{12}{7}\)

Số hơi to??

\(b,-12:\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(=-12:\left(\frac{9}{12}-\frac{10}{12}\right)^2\)

\(=-12:\left(-\frac{1}{12}\right)^2\)

\(=-12\cdot144\)

\(=-1728\)

\(a)=\left(\frac{11}{24}+\frac{13}{24}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)+\frac{1}{2}\)

\(=\frac{24}{24}-\frac{41}{41}+\frac{1}{2}\)

\(=1-1+\frac{1}{2}\)

\(=\frac{1}{2}\)

a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)

b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)

\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)

c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)

\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)

d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)

cả 2 cái cộng lại hay là từng cái một vậy bạn?

a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?

\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)

b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)

=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)

=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)