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, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)

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

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

Câu 2: n= 12

Do A=\(\frac{\left(2x2\right)^6x\left(2x3\right)^6}{3^6x2^6}=2^{12}\)

Bạn có thể giả thích rõ hơn ko???

Bài 5: GTNN chứ nhỉ?

Với mọi gt của \(x;y\in R\) ta có:

\(x^2+3\left|y-2\right|+1\ge1\)

Hay \(A\ge1\) với mọi gt của \(x;y\in R\)

Dấu "=" sảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

Vậy..................

Bài 6: GTLN chứ?

Với mọi giá trị của \(x\in R\) ta có:

\(-\left(2x-1\right)^2\le0\Rightarrow-5-\left(2x-1\right)^2\le-5\)

Hay \(B\le5\) với mọi giá trị của \(x\in R\)

Dấu "=" sảy ra khi và chỉ khi \(x=\dfrac{1}{2}\)

Vậy...................

Bài 4 :

\(a,3^{15}-9^6=3^{15}-\left(3^2\right)^6=3^{15}-3^{12}=3^{12}\left(3^3-1\right)=3^{12}.26=3^{12}.2.13⋮\left(đpcm\right)\)

\(b,8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}=2^{18}\left(2^3-1\right)=2^{18}.7=2^{17}.2.7=2^{17}.14⋮14\left(đpcm\right)\)

Bài 5 :

\(A=1^2+3^2+6^2+9^2+.............+39^2\)

\(=1+3^2+\left(6^2+9^2+.........+39^2\right)\)

\(=10+3^2\left(2^2+3^2+.........+13^2\right)\)

\(=10+3^2.818\)

\(=10+9.818\)

\(=7372\)

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

c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)

\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)

d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)

e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)

i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)

\(=3^{40}-1\)