Đáp án:
a) x = $\frac{1}{4}$
b) x = 0
c) x =2
d) x = $\frac{1}{2}$
e) x = - $\frac{8}{13}$
Giải thích các bước giải:
a) $\sqrt{x² -4x + 1}$ = x
⇔ $\left \{ {{x\geq0} \atop {{x² -4x + 1} = x²}} \right.$
⇔ $\left \{ {{x\geq0} \atop {{-4x + 1} = 0}} \right.$
⇔ $\left \{ {{x\geq0} \atop {{x } =\frac{1}{4} }} \right.$ (TMĐK)
b) $\sqrt{x² +x + 1}$ = x +1
⇔ $\left \{ {{x +1\geq0} \atop {{x² +x + 1} = (x+1)²}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{x² +x + 1} = x² + 2x +1}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{x² +x + 1 - x² - 2x -1} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{-x} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{x} = 0}} \right.$ (TMĐK)
c) $\sqrt{4x² -8x + 1}$ = x -1
⇔ $\left \{ {{x -1\geq0} \atop {{4x² -8x + 1} = (x-1)²}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{4x² -8x + 1} = x² - 2x +1}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{4x² -8x + 1 - x² + 2x -1} = 0}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{3x² -6x } = 0}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{3x . (x -2) } = 0}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{\left[ \begin{array}{l}3x=0\\x-2=0\end{array} \right. }}} \right.$
⇔ $\left \{ {{x \geq1} \atop {{\left[ \begin{array}{l}x=0 (Loại)\\x=2 (TMĐK)\end{array} \right. }}} \right.$
d)$\sqrt{5x² -2x + 2}$ = x +1
⇔ $\left \{ {{x +1\geq0} \atop {{5x² -2x + 2} = (x+1)²}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{5x² -2x +2} = x² + 2x +1}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{5x² -2x +2 - x² - 2x -1} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{4x² -4x + 1} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{(2x - 1)²} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{2x - 1} = 0}} \right.$
⇔ $\left \{ {{x \geq-1} \atop {{x} =\frac{1}{2} }} \right.$ (TMĐK)
e) $\sqrt{4x² -x + 1}$ - 2x = 3
⇔ $\sqrt{4x² -x + 1}$ = 3 + 2x
⇔ $\left \{ {{3-2x\geq0} \atop {{4x² -x + 1} = (3+2x)²}} \right.$
⇔ $\left \{ {{x\leq\frac{3}{2}} \atop {{4x² -x + 1} = 9+2.2x.3+ 4x²}} \right.$
⇔ $\left \{ {{x\leq\frac{3}{2}} \atop {{4x² -x + 1 - 9-12x- 4x²} = 0}} \right.$
⇔ $\left \{ {{x\leq\frac{3}{2}} \atop {{-13x - 8} = 0}} \right.$
⇔ $\left \{ {{x\leq\frac{3}{2}} \atop {{x } = -\frac{8}{13}}} \right.$ (TMĐK)
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@Phương
