a) -5+|3x-1|+6=|-4| b) $(x-1)^{2}$= $(x-1)^{4}$

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1. Đáp án :

   `a)x∈{-2/3; 4/3}`

   `b)x∈{0;1;2}`

   Giải thích các bước giải :

   `a)-5+|3x-1|+6=|-4|`

   `<=>|3x-1|+6-5=4`

   `<=>|3x-1|=4-1`

   `<=>|3x-1|=3`

   `<=>`\(\left[ \begin{array}{l}3x-1=3\\3x-1=-3\end{array} \right.\)

   `<=>`\(\left[ \begin{array}{l}3x=4\\3x=-2\end{array} \right.\)

   `<=>`\(\left[ \begin{array}{l}x=\frac{4}{3}\\3x=\frac{-2}{3}\end{array} \right.\)

   Vậy `x∈{-2/3; 4/3}`

   `b)(x-1)^2=(x-1)^4`

   `<=>(x-1)^2-(x-1)^4=0`

   `<=>(x-1)^2×[1-(x-1)^2]=0`

   `<=>(x-1)^2(1-x+1)(1+x-1)=0`

   `<=>x(2-x)(x-1)^2=0`

   `<=>`\(\left[ \begin{array}{l}x=0\\2-x=0\\(x-1)^2=0\end{array} \right.\)

   `<=>`\(\left[ \begin{array}{l}x=0\\x=2\\x-1=0\end{array} \right.\)

   `<=>`(\left[ \begin{array}{l}x=0\\x=2\\x=1\end{array} \right.\)

   Vậy `x∈{0;1;2}`

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2. a) `-5+|3x-1|+6=|-4|`

   `-5+|3x-1|+6=4`

   `|3x-1|=3`

   TH1: `3x-1=3`

   `3x=4`

   `x=4/3`

   TH2: `3x-1=-3`

   `3x=-2`

   `x=-2/3`

   Vậy `x=4/3 ; x=-2/3`

   b) `(x-1)^2=(x-1)^4`

   `(x-1)=(x-1)^2`

   TH1: `x-1 = (x-1)^2`

   ` (x-1)^2-(x-1)=0`

   `(x-1)(x-1-1)=0`

   `(x-1)(x-2)=0`

   `x=1 \vee x=2`

   TH2: `-(x-1)=(x-1)^2`

   `(x-1)^2+(x-1)=0`

   `(x-1)(x-1+1)=0`

   `x(x-1)=0`

   `x = 0 \vee x=1`

   Vậy `x=0 ; x=1 ; x=2`

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