64⁸=(2⁶)⁸=2⁴⁸

16¹²=(2⁴)¹²=2⁴⁸

Vì 2⁴⁸=2⁴⁸ nên 64⁸=16¹²

64⁸ = (2⁶)⁸ = 2⁴⁸

16^{12 =} (2⁴)¹² = 2⁴⁸

Vì 2⁴⁸ = 2⁴⁸ => 16¹² = 64⁸

Vậy 64⁸ = 16¹²

64² = (2⁶)² = 2¹²

16¹² = (2⁴)¹² = 2⁴⁸

Vì 2¹² < 2⁴⁸ => 64² < 16¹²

Vậy 64² < 16¹²

clgt???

Đặt B =\(32^{17}.8^{19}\)

So sánh :\(A=64^{11}.16^{11}\)và  \(B=32^{17}.8^{19}\)

TA có :\(A=64^{11}.16^{11}=\left(64.16\right)^{11}=1024^{11}=\left(2^{10}\right)^{11}\)\(=2^{110}\)

            \(B=32^{17}.8^{19}=\left(2^5\right)^{17}.\left(2^3\right)^{19}=2^{85}.2^{57}\)\(=2^{142}\)

 VÌ  A < B ( 2¹¹⁰< 2¹⁴²)

 Nên A < 32¹⁷.8¹⁹

\(6^8và16^{12}=\left(6.8\right)^0và\left(16.3\right)^9=48< 48^9\)

6⁸ = (6²)⁴ = 36⁴

16¹² = (16³)⁴ = 4096⁴

Do 36 < 4096 nên 36⁴ < 4096⁴

Vậy 6⁸ < 16¹²

Cách 2:

Do 6 < 16 nên 6¹² < 16¹²   (1)

Do 8 < 12 nên 6⁸ < 6¹²    (2)

Từ (1) và (2) ⇒ 6⁸ < 16¹²

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

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co don quaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

\(8^3.64^2=8^3.8^4=8^7>8^5\)