Đáp án + Giải thích các bước giải:
` 2) `
`a)` `(11x - 3/4)^3 + 21 9/17 = 29 9/17`
` (11x - 3/4)^3 = 29 9/17 - 21 9/17`
` (11x - 3/4)^3 = (29 - 21) + (9/17 - 9/17)`
`(11x - 3/4)^3 = 8 + 0`
`(11x - 3/4)^3 = 8`
`=>` `11x - 3/4 = 2`
`11x = 2 + 3/4`
` 11x = 8/4 + 3/4`
`11x = 11/4`
` x = 11/4 : 11`
`x = 11/4 . 1/11`
`x = 1/4`
Vậy `x = 1/4`
`b)` `(3x - 7/3)^2 - 5/4 = -1`
`(3x - 7/3)^2 = -1 + 5/4`
`(3x - 7/3)^2 = (-4)/4 + 5/4`
` (3x - 7/3)^2 = 1/4`
`=>` \(\left[ \begin{array}{l}3x - \dfrac{7}{3} = \dfrac{1}{2} \\3x - \dfrac{7}{3} = \dfrac{-1}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x = \dfrac{1}{2} + \dfrac{7}{3}\\3x = \dfrac{-1}{2} + \dfrac{7}{3}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x = \dfrac{3}{6} + \dfrac{14}{6}\\3x =\dfrac{-3}{6} + \dfrac{14}{6} \end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x = \dfrac{17}{6}\\3x = \dfrac{11}{6}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{17}{6} : 3\\x=\dfrac{11}{6} : 3\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{17}{18} \\x=\dfrac{11}{18}\end{array} \right.\)
Vậy `x in {17/18; 11/18}`
`c)` `17/2 - |2x - 3/4| = -7/4`
` |2x - 3/4| = 17/2 - (-7)/4`
`|2x - 3/4| = 34/4 + 7/4`
`|2x - 3/4| = 41/4`
`=>` \(\left[ \begin{array}{l}2x - \dfrac{3}{4} = \dfrac{41}{4}\\2x - \dfrac{3}{4} = \dfrac{-41}{4}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x = 11\\2x = \dfrac{-19}{2}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{11}{2}\\x=\dfrac{-19}{4} \end{array} \right.\)
Vậy `x in {11/2; (-19)/4}`
`d)` `|x| + x = 2/3`
Vì `|x| ≥ 0` với `∀x`
`=>` `x ≥ 0`
`=>` `|x| + x = 2/3`
`x + x = 2/3`
`2x = 2/3`
`x = 2/3 : 2`
`x = 2/3 . 1/2`
`x = 1/3`
Vậy `x = 1/3`
`e)` `2^x . (2^3)^2 = 1024`
`2^x . 2^6 = 1024`
`2^x . 2^6 = 2^10`
`2^x = 2^10 : 2^6`
`2^x = 2^4`
`=>` `x = 4`
Vậy `x = 4`
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