Đáp án + Giải thích các bước giải:

 `1)` 

`a)` `2^5. 8^4`

`= 2^5 . (2^3)^4`

`= 2^5 . 2^12`

`= 2^17`

`b)` `25^6 : 125^3`

`= (5^2)^6 . (5^3)^3`

`= 5^12 . 5^9`

`= 5^21`

`c)` `625^5 : 25^7`

`= (5^4)^5 : (5^2)^7`

`= 5^20 : 5^14`

`= 5^6`

`d)` `12^3 : 3^3`

`= (12:3)^3`

`= 4^3`

`2)` Tìm `x in NN,` biết:

`a)` `2^x - 15 = 17`

   `2^x = 17 + 15`

   `2^x = 32`

   `2^x = 2^5`

`=>` `x = 5`

`b)` `(2x + 1)^3 = 9 . 81`

    `(2x + 1)^3 = 729`

    `(2x + 1)^3 = 9^3`

`=>` `2x + 1 = 9`

     `2x = 9 - 1`

    `2x  = 8`

    ` x = 8 : 2`

   ` x = 4`

`c)` `(2x - 3)^2 - 9 = 16`

`   (2x - 3)^2 = 16 + 9`

  ` (2x - 3)^2 = 25`

  ` (2x - 3)^2 = 5^2`

`=>` `2x - 3 = 5`

      `2x = 5 + 3`

     `2x = 8`

       `x = 8 : 2`

       `x = 4`

`d)` `5^(2x + 3) - 2 . 5^2 = 5^2 . 3`

   `5^(2x + 3) - 2 . 25 = 25 . 3`

  `5^(2x + 3) - 50 = 75`

  `5^(2x + 3) = 75 + 50`

  `5^(2x + 3) = 125`

  `5^(2x + 3) = 5^3`

`=>` `2x +3 = 3`

      `2x = 3 -  3`

      `2x = 0 `

       `x = 0 : 2`

       `x = 0`

`3)` So sánh:

`a)` `27^11` và `81^8`

Ta có: `27^11 = (3^3)^11 = 3^33`

         `81^8 = (3^4)^8 = 3^32`

Vì `3^33 > 3^32 => 27^11 > 81^8`

`b)` `625^5` và `125^7`

Ta có: `625^5 = (5^4)^5 = 5^20`

         `125^7 = (5^3)^7 = 5^21`

Vì `5^20 < 5^21 => 625^5 < 125^7`

`c)` `5^23` và `6 . 5^22`

Ta có: `6. 5^22 > 5 . 5^22 = 5^23`

`=>` `5^23 < 6 . 5^22`

`d)` `16^9` và `8^24`

Ta có: `16^9 = (2^4)^9 = 2^36`

          `8^24 = (2^3)^24 = 2^72`

Vì `2^36 < 2^72 => 16^9 < 8^24`

`4)` Tính tổng:

`a)` `S = 1 + 2 + 2^2 + 2^3 + ... + 2^99 + 2^100`

`=>` `2S = 2 + 2^2 + 2^3 + ... + 2^100 + 2^101`

`=>` `\underbrace{2S-S}_{S} = (2 + 2^2 + 2^3 + ... + 2^100 + 2^101) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^99 + 2^100) `

`=>` `S = 2^101 - 1`

`b)` `S = 1 + 3 + 3^2 + 3^3 + ... + 3^99 + 3^100`

`=>` `3S = 3 + 3^2 + 3^3 + ... + 3^100 + 3^101`

`=>`  `\underbrace{3S-S}_{2S} = (3 + 3^2 + 3^3 + ... + 3^100 + 3^101) - ( 1 + 3 + 3^2 + 3^3 + ... + 3^99 + 3^100) `

`=>` `2S = 3^101 - 1`

`=>` `S = (3^101 - 1)/2`

`5)` Tìm số tự nhiên `n,` biết:

`a)` `9 < 3^n < 81`

`=>` `3^2 < 3^n < 3^4`

`=>` `n = 3`

`b)` `25 ≤ 5^n ≤ 125`

`=>` `5^2 ≤ 5^n ≤ 5^3`

`=>` `n in {2; 3}`

CHÚC BẠN HỌC TỐT!

Bài 1 :

$a) 2^5.8^4=2^5.2^{12}=2^{17}$

$b) 25^6.125^3=5^{12}.5^{9}=5^{21}$

$c)  625^5:25^{7}=25^{10}:25^{7}=25^{3}=5^6$

$d) 12^3:3^3=(12:3)^3=4^3$

Bài 2:

$a) 2^{x}-15=17 ⇔ 2^x=32 ⇔ 2^x=2^5 ⇔ x=5$

$b) (2x+1)^3=9.81⇔(2x+1)^3=9^3⇔2x+1=9 ⇔ x=4$

$c) (2x-3)^2-9=16 ⇔ (2x-3)^2=25 ⇔ 2x-3=5 ⇔ 2x=8 ⇔ x=4$

$d) 5^{2x+3}-2.5^2=5^2.3$

$⇔5^{2x}+125-50=75$

$⇔5^{2x}=0$

$⇔x=0$

Bài 3:

$a) 27^{11}=(3^3)^{11}=3^{33}$

$81^8=(3^4)^8=3^{32}$

$⇒27^{11}>81^8$

$b) 625^5=(5^4)^5=5^{20}$

$125^7=(5^3)^7=5^{21}$

$⇒625^5<125^7$

$c) 6.5^{22} > 5.5^{22} = 5^{23}$

$⇒6.5^{22}>5^{23}$

$d) 16^9=2^{36}$

$8^{25}=2^{75}$

$⇒16^9<8^{25}$

Bài 4:

$a) S=1+2+2^2+...+2^{99}+2^{100}$

$⇒2S=2+2^2+2^3+...+2^{101}$

$⇒2S-S=(2+2^2+2^3+...+2^{101})-1+2+2^2+...+2^{99}+2^{100}$

$⇒S=2^{101}-1$

$b) S=1+3+3^2+...+3^{100}$

$⇒3S=3+3^2+...+3^{101}$

$⇒3S-S=(3+3^2+...+3^{101})-(1+3+...+3^{100})$

$⇒2S=3^{101}-1 ⇒ S =\dfrac{3^{101}-1}{2}$

Bài 5:

$a) 9<3^n<81$

$⇔3^2<3^n<3^4$

$⇔n=3$

$b) 25 \leq 5^n \leq 125$

$⇔5^2 \leq 5^n \leq 5^3$

$⇔n ∈ \{ 2 ; 3 \}$