Đáp án:
a, `x \in {9 ; -9}`
b, `x = 6`
c, `x = 5`
d, `x = 7`
d, `x = 4`
e, ` x = 7`
f, `x \in {1 ; 0}`
g, `x \in {7 ; 8}`
Giải thích các bước giải:
Bài `4`:
a, `x^2 = 2^3 + 3^2 + 4^3`
` x^2 = 8 + 9 + 64`
` x^2 = 81`
`=> x = \pm 9`
Vậy `x \in {9 ; -9}`
b, `x^3 = 216`
` x^3 = 6^3`
`=> x = 6`
Vậy `x = 6`
c, `2^x - 15 = 17`
` 2^x = 17 + 15`
` 2^x = 32`
` 2^x = 2^5`
`=> x = 5`
Vậy `x = 5`
d, `2^x + 4 = 132`
` 2^x = 132 - 4`
` 2^x = 128`
` 2^x = 2^7`
`=> x = 7`
Vậy `x = 7`
d, `2 . 3^x = 162`
` 3^x = 162 : 2`
` 3^x = 81`
` 3^x = 3^4`
`=> x = 4`
Vậy `x = 4`
e, `3 . 2^x = 384`
` 2^x = 384 : 3`
` 2^x = 128`
` 2^x = 2^7`
`=> x = 7`
Vậy `x = 7`
f, `x^3 = x^2`
`=> x^3 - x^2 = 0`
`x^2(x - 1) = 0`
⇒\(\left[ \begin{array}{l}x^2 = 0\\x - 1= 0 \end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 0\\x= 1 \end{array} \right.\)
Vậy `x \in {0 ; 1}`
g, `(2x - 15)^5 = (2x - 15)^3`
`=> (2x - 15)^5 - (2x - 15)^3 = 0`
`=> (2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
⇒\(\left[ \begin{array}{l}(2x - 15)^3 = 0\\ (2x - 15)^2 - 1 = 0 \end{array} \right.\)
⇒\(\left[ \begin{array}{l}2x - 15 = 0\\ (2x - 15)^2 = 1 \end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = \frac{15}{2}\\ 2x - 15 = 1 \\ 2x - 15 = -1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = \frac{15}{2} (KTM đk: x ∈ N)\\ x = 8 \\ x = 7\end{array} \right.\)
Vậy `x \in {7 ; 8 }`
