a)0,3¹⁰=(0,3²)¹⁰=0,9¹⁰

0,9>0,1=>0,9¹⁰>0,1¹⁰hay0,3²⁰>0,1¹⁰

b)4³⁰+3²⁰=(4³)¹⁰+(3²)¹⁰=(4³+3²)¹⁰=73¹⁰

3.24=72

có 73>72 => 73¹⁰>72¹⁰ hay 4³⁰+3²⁰>3.24¹⁰ (câu này mình ko chắc đúng đâu)

Ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)

\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)

Vì \(2^{40}< 2^{50}\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)hay \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(0,3\right)^{20}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)

Vì \(0,09< 0,1\Rightarrow\left(0,09\right)^{10}< \left(0,1\right)^{100}\)

hay \(\left(0,3\right)^{20}< \left(0,1\right)^{10}\)

Ta có: \(A=\frac{7^{10}}{1+7+7^2+...+7^9}\)

\(\Rightarrow\frac{1}{A}=\frac{1+7+7^2+...+7^9}{7^{10}}=\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7}\)

Lại có: \(B=\frac{5^{10}}{1+5+5^2+...+5^9}\)

\(\Rightarrow\frac{1}{B}=\frac{1+5+5^2+...+5^9}{5^{10}}=\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}\)

Ta có: \(7^{10}>5^{10}\Rightarrow\frac{1}{7^{10}}< \frac{1}{5^{10}}\)

         \(7^9>5^9\Rightarrow\frac{1}{7^9}< \frac{1}{5^9}\)

         \(7^8>5^8\Rightarrow\frac{1}{7^8}< \frac{1}{5^8}\)

          \(...............................\)

         \(7>5\Rightarrow\frac{1}{7}< \frac{1}{5}\)

\(\Rightarrow\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7}< \frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}\)

\(\Rightarrow\frac{1}{A}< \frac{1}{B}\Rightarrow A>B\)

Chúc bạn học tốt !!!

     Giải :

\(B=\left[\left(-\frac{3}{7}\right)^5\right]^4\)

\(B=\left(-\frac{3}{7}\right)^{20}\)

\(A=\frac{3}{7}\cdot\left(\frac{3}{7}\right)^{19}\)

\(A=\left(\frac{3}{7}\right)^{20}\)

\(\Rightarrow A>B\)

           [ hoq chắc ]

ta có: 5^10+6^10=(5+6)^10=11^10>7^10

=>5^10+6^10>7^10

mình nhầm: 5^10+6^10<7^10

a) 10²⁰ = 100¹⁰  => 10²⁰ > 90¹⁰

b) 0,3²⁰ = 0,9¹⁰  => 0,1¹⁰ < 0,9¹⁰

a) 27¹¹ và 84⁸

    27¹¹ > 84⁸

b) 625⁵ và 125⁷

    625⁵ > 125⁷

c) 5²⁵ và 6*5²²

    5²⁵ > 6*5²²

đ) 7*2¹³ và 2¹⁶

    7*2^{13 <} 2¹⁶

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30¹² = (30⁴)³ = 810000³

10¹⁸ = (10⁶)³ = 1000000³

Ta thấy 810000³ < 1000000³ nên 30¹² < 10¹⁸

5³⁶ = (5³)¹² = 125¹²

10²⁴ = (10²)¹² = 100¹²

Ta thấy 125¹² > 100¹² nên 5³⁶ > 10²⁴

1) Ta có:

8⁷ - 2¹⁸

= (2³)⁷ - 2¹⁸

= 2²¹.2¹⁸

= 2¹⁸.(2³ - 1)

= 2¹⁷.2.7

= 2¹⁷.14 chia hết cho 14 (đpcm)

2) Ta có:

2⁹¹ = (2¹³)⁷ = 8192⁷

5³⁵ = (5⁵)⁷ = 3125⁷

Vì 8192⁷ > 3125⁷

=> 2⁹¹ > 5³⁵