Đáp án + Giải thích các bước giải:
Bài `22`:
a, `(2x + 1)^3 = 9 . 81`
`(2x + 1)^3 = 9 . 9^2`
` (2x + 1)^3 = 9^3`
`=> 2x + 1 = 9`
`=> 2x = 9 - 1`
`=> 2x = 8`
`=> x = 4`
Vậy `x = 4`
b, `5^x + 5^(x + 2) = 650`
`5^x . (1 + 25) = 650`
` 26 . 5^x = 650`
` 5^x = 650 : 26`
` 5^x = 25`
` 5^x = 5^2`
`=> x = 2`
Vậy `x = 2`
c, `(4x - 1)^2 = 25 . 9`
` (4x - 1)^2 = (5 . 3)^2`
` (4x - 1)^2 = 15^2 = (-15)^2`
`=> 4x - 1 = 15 ` hoặc `4x - 1 = -15`
Nếu `4x - 1 = 15` ; Nếu `4x - 1 = -15`
`=> 4x = 16` ; ` => 4x = -14`
`=> x = 4` ; ` => x = -14/4`
Vậy `x \in {4 ; -14/4}`
Bài `23`:
a, `2^x + 2^(x + 3) = 144`
` 2^x . (1 + 8) = 144`
` 2^x . 9 = 144`
`=> 2^x = 16`
`=> 2^x = 2^4`
`=> x = 4`
Vậy `x = 4`
b, `3^(2x + 2) = 9^(x + 3)`
` 3^(2x + 2) = (3^2)^(x + 3)`
` 3^(2x + 2) = 3^(2x + 6)`
`=> 2x + 2 = 2x + 6`
`=> 0x = 4` (Vô lý)
Vậy `x \ in ∅`
c, `(x - 5)^4 = (x - 5)^6 (x \ge 5)`
`=> (x - 5)^6 - (x - 5)^4 = 0`
` (x - 5)^4 . ( (x - 5)^2 - 1) = 0`
`=> (x - 5)^4 = 0` hoặc `(x - 5)^2 - 1 = 0`
Nếu `(x - 5)^4 = 0` ; Nếu `(x - 5)^2 -1 = 0`
`=> x - 5 = 0` ; `(x - 5)^2 = 1`
`=> x = 5 (TM)` ; \(\left[ \begin{array}{l}x - 5 = 1\\x - 5 = -1\end{array} \right.\)
; \(\left[ \begin{array}{l}x = 6 (TM)\\x = 4(KTM)\end{array} \right.\)
Vậy `x \in {5 ; 6}`
d, `x^15 = x^2`
` x^15 - x^2 = 0`
` x^2(x^13 - 1) = 0`
`=> x^2 = 0` hoặc `x^13 - 1 = 0`
Nếu `x^2 = 0` ; Nếu `x^13 - 1 = 0`
`=> x = 0` ; `x^13 = 1 => x = 1`
Vậy `x \in {0 ; 1}`
