Đáp án + Giải thích các bưdớc giải:
`16)` `1/9. x^2 - 2x = 0`
`1/9 . x . x - 2x = 0`
`x. (1/9. x - 2) = 0`
`=>` \(\left[ \begin{array}{l}x=0\\\dfrac{1}{9}x - 2=0\end{array} \right.\)
Ta có: `1/9. x - 2 = 0`
`1/9. x = 0 + 2`
`1/9 . x = 2`
`x = 2 : 1/9`
`x = 2 . 9`
`x = 18`
Vậy `x in {0; 18}`
`22)` `7^(x + 2) + 2 . 7^x = 357`
` 7^x . 7^2 + 2. 7^x = 357`
` 7^x . (7^2 + 2) = 357`
`7^x . (49 + 2) = 357`
` 7^x . 51 = 357`
` 7^x = 357 : 51`
` 7^x = 7`
`=>` `x = 1`
`23)` `x^5 = x^3`
`=>` `x^5 - x^3 = 0`
`x^3 . (x^2 - 1) = 0`
`=>` \(\left[ \begin{array}{l}x^3 = 0\\x^2 - 1 = 0\end{array} \right.\)
`+)` `x^3 = 0`
`=>` `x = 0`
`+)` `x^2 - 1 = 0`
`x^2 = 0 + 1`
`x^2 = 1`
`=>` `x = ± 1`
Vậy `x in {0; ±1}`
`24)` `x^6 - x^4 = 0`
`x^4. (x^2 - 1) = 0`
`=>` \(\left[ \begin{array}{l}x^4 = 0\\x^2 - 1 = 0\end{array} \right.\)
`+)` `x^4 = 0`
`=>` `x = 0`
`+)` `x^2 - 1 = 0`
` x^2 = 0 + 1`
`x^2 = 1`
`=>` `x = ± 1`
Vậy `x in {0; ±1}`
`25)` `x^5 + x^2 = 0 `
`x^2. (x^3 + 1) = 0`
`=>` \(\left[ \begin{array}{l}x^2 = 0\\x^3 + 1 = 0\end{array} \right.\)
`+)` `x^2 = 0`
`=>` `x = 0`
`+)` `x^3 + 1 = 0`
`x^3 = 0 - 1`
`x^3 = -1`
`=>` `x = -1`
Vậy `x in {0; -1}`
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