`{2- x}/4= {3x- 1}/-3`
`{12- 3x}/32= {6}/{4- x}`
`{2- x}/4= {3x- 1}/-3` `{12- 3x}/32= {6}/{4- x}`
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`{2- x}/4= {3x- 1}/-3`
`{12- 3x}/32= {6}/{4- x}`
Neala
Các bước giải:
`(2 -x)/4 = (3x -1)/-3`
`⇒ -3.(2 -x) = 4.(3x -1)`
`⇒ -6 +3x = 12x -4`
`⇒ -9x = 6`
`⇒ x = -2/3`
Vậy `x = -2/3`
——————
`(12 -3x)/32 = 6/(4 -x)` `(ĐKXĐ:` `x ne 4)`
`⇒ (12 -3x).(4 -x) = 32.6`
`⇒ 48 -24x +3x² = 192`
`⇒ 3x² -24x -144 = 0`
`⇒ 3x.(x -12) +12.(x -12) = 0`
`⇒ (x -12).(3x +12) = 0`
$⇒ \left[ \begin{array}{l}x -12=0\\3x +12=0\end{array} \right. ⇒ \left[ \begin{array}{l}x=12\\x=-4\end{array} \right.$ $(T/m$ `đkxđ)`
Vậy `x ∈ {12; -4}`
Bơ
`(2-x)/4 = (3x-1)/(-3)`
`<=> -3(2-x) = 4(3x-1)`
`<=> -6+3x=12x-4`
`<=> -4x = 2`
`<=> x = -1/2`
.
`(12-3x)/32 = 6/(4-x)`
`<=> 32.6 = (12-3x)(4-x)`
`<=> 192 = 3(4-x)(4-x)`
`<=> 64 = (4-x)^2`
`<=> (x-4)^2 = 8^2`
`<=>` \(\left[ \begin{array}{l}x=12\\x=-4\end{array} \right.\)