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a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)

a) Ta có: \(\frac{7}{8}+\frac{15}{23}+\frac{1}{8}+\left(-\frac{15}{23}\right)+\frac{1}{3}\)
\(=\left(\frac{7}{8}+\frac{1}{8}\right)+\left(\frac{15}{23}-\frac{15}{23}\right)+\frac{1}{3}\)
\(=1+\frac{1}{3}=\frac{4}{3}\)
b) Ta có: \(\frac{3-\frac{2}{3}}{\frac{1}{3}-3}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{1}{\frac{1}{3}-3}=\left(3-\frac{2}{3}\right)\cdot\frac{1}{-\frac{8}{3}}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{-3}{8}\)
\(=-\frac{9}{8}+\frac{1}{4}=\frac{-9}{8}+\frac{2}{8}=-\frac{7}{8}\)
c) Ta có: \(8\cdot\left(-\frac{1}{2}\right)^3+3\cdot\left(-\frac{1}{2}\right)^2-0,75\)
\(=8\cdot\frac{-1}{8}+3\cdot\frac{1}{4}-\frac{3}{4}\)
\(=-1\)

a) \(-\frac{4}{15}\cdot\left(-\frac{3}{8}\right)\cdot\left(\frac{15}{-4}\right)\cdot8\)
\(=\left(-\frac{4}{15}\cdot\left(\frac{15}{-4}\right)\right)\cdot\left(-\frac{3}{8}\cdot8\right)\)
\(=1\cdot\left(-3\right)=-3\)
b) \(-\frac{2}{3}\cdot\frac{4}{19}+\frac{5}{3}\cdot\frac{4}{19}\)
\(=\frac{4}{19}\cdot\left(\frac{5}{3}-\frac{2}{3}\right)\)
\(=\frac{4}{19}\cdot1=\frac{4}{19}\)
c) \(-\frac{1}{2}\cdot\frac{5 }{3}+\frac{1}{2}\cdot\frac{8}{3}\)
\(=\frac{1}{2}\left(\frac{8}{3}-\frac{5}{3}\right)\)
\(=\frac{1}{2}\cdot1=\frac{1}{2}\)

Câu a : - 891 / 100 = - 8 , 91
Câu b : - 21 / 4 = - 5 , 25
Câu c : 5 / 2 = 2 , 5
Câu d : 4096 / 387 = 11 ( đã làm tròn )
Câu e : - 3 / 2 = - 1 , 5

a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)

1)
Ta có: \(\left(x-\frac{1}{2}\right)^3=8\Rightarrow\left(x-\frac{1}{2}\right)^3=2^3\)
\(\Rightarrow x-\frac{1}{2}=2\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}\)
\(\left(x-1\right)^3=\frac{8}{27}\Rightarrow\left(x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow x-1=\frac{2}{3}\Rightarrow x=\frac{2}{3}+1=\frac{5}{3}\)

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
\(=\left(9-1-1-...1\right)+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)
\(=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}=\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
\(=10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)=10B\)
vậy A:B=10
( 2 . * - 1 ) ^ 3 = -8
( 2 . * - 1 ) ^ 3 = ( - 2 ) ^ 3
=> 2 . * - 1 = -2
2 . * = -1
* = - 0 , 5