2 câu trả lời
Đáp án+Giải thích các bước giải:
`\sqrt{x-1}-\sqrt{x-2}=\sqrt{x-3}(x≥3)`
`⇔x-1-2\sqrt{(x-1)(x-2)}+x-2=x-3`
`⇔2x-3-x+3=2\sqrt{x^2-3x+2}`
`⇔x=2\sqrt{x^2-3x+2}`
`⇔x^2=4(x^2-3x+2)`
`⇔4x^2-12x+8-x^2=0`
`⇔3x^2-12x+8=0`
`⇔(\sqrt{3}x)^2-2.\sqrt{3}x.2\sqrt{3}+12-4=0`
`⇔(\sqrt{3}x-2\sqrt{3})^2=4`
$⇔\left[\begin{matrix}\sqrt{3}x-2\sqrt{3}=2\\\sqrt{3}x-2\sqrt{3}=-2\end{matrix}\right.$
$⇔\left[\begin{matrix}\sqrt{3}x=2+2\sqrt{3}\\\sqrt{3}x=-2+2\sqrt{3}\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{6+2\sqrt{3}}{3}\\x=\dfrac{6-2\sqrt{3}}{3}(L)\end{matrix}\right.$
Vậy `S={\frac{6+2\sqrt{3}}{3}}`
Đáp án: $x=\dfrac{6+2\sqrt{3}}{3}$
Giải thích các bước giải:
$\sqrt{x-1}-\sqrt{x-2}=\sqrt{x-3}$ (ĐK: $x\ge 3$)
$\Leftrightarrow \sqrt{x-1}=\sqrt{x-2}+\sqrt{x-3}$
$\Leftrightarrow x-1=x-2+x-3+2\sqrt{\left( x-2 \right)\left( x-3 \right)}$
$\Leftrightarrow 4-x=2\sqrt{{{x}^{2}}-5x+6}$
$\Leftrightarrow {{\left( 4-x \right)}^{2}}=4\left( {{x}^{2}}-5x+6 \right)$ ($3\le x\le 4$)
$\Leftrightarrow 16-8x+{{x}^{2}}=4{{x}^{2}}-20x+24$
$\Leftrightarrow 3{{x}^{2}}-12x+8=0$
$\Leftrightarrow x=\dfrac{6+2\sqrt{3}}{3}$ (nhận) hoặc $x=\dfrac{6-2\sqrt{3}}{3}$ (loại)