Đáp án:
Giải thích các bước giải:
a) `|2x^2-7x+6|=x-2`
TH1: `x \in (-∞;3/2]∪[2;+∞)`
`2x^2-7x+6=x-2`
`⇔ 2x^2-8x+8=0`
`⇔ (x-2)^2=0`
`⇔ x=2\ (TM)`
TH2: `x \in (3/2;2)`
`2x^2-7x+6=2-x`
`⇔ 2x^2-6x+4=0`
`⇔` \(\left[ \begin{array}{l}x=2\ (L)\\x=1\ (L)\end{array} \right.\)
Vậy `S={2}`
b) `|4x^2-6x+2|=x+1`
TH1: `x \in (-∞;1/2]∪[1;+∞)`
`4x^2-6x+2=x+1`
`⇔ 4x^2-7x+1=0`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{7+\sqrt{33}}{8}\ (TM)\\x=\dfrac{7-\sqrt{33}}{8}\ (TM)\end{array} \right.\)
TH2: `x \in (1/2;1)`
`4x^2-6x+2=-x-1`
`⇔ 4x^2-5x+3=0` (VN do `\Delta < 0)`
Vậy `S={\frac{7+\sqrt{33}}{8};\frac{7-\sqrt{33}}{8}}`
c) `|x^2-2x|=x`
TH1: `x \in (-∞;0]∪[2;+∞)`
`x^2-2x=x`
`⇔ x^2-3x=0`
`⇔ x(x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\ (TM)\\x=3\ (TM)\end{array} \right.\)
TH2: `x \in (0;2)`
`x^2-2x=-x`
`⇔ x^2-x=0`
`⇔` \(\left[ \begin{array}{l}x=0\ (L)\\x=1\ (TM)\end{array} \right.\)
Vậy `S={0;3;1}`
