$A=\sqrt[]{(1-3x)^2)}+\sqrt[]{3x-2)^2}$
$=|1-3x|+|3x-2|$
Áp dụng bất đẳng thức $|a|+|b|≥|a+b|$; dấu = xảy ra $⇔a.b≥0$ ta được:
$|1-3x|+|3x-2|≥|1-3x+3x-2|=1$
Hay $A≥1$
Dấu $=$ xảy ra $⇔$\(\left[ \begin{array}{l}\begin{cases}1-3x \geq 0\\3x-2 \geq 0\end{cases}\\\begin{cases}1-3x <0 \\3x-2 < 0\end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\begin{cases}x ≤\dfrac{1}{3}\x ≥\dfrac{3}{2}\end{cases}⇒\text{Loại}\\\begin{cases}x>\dfrac{1}{3} \\x>\dfrac{2}{3}\end{cases}⇒x>\dfrac{2}{3}\end{array} \right.\)
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2,Ta có: $\dfrac{4}{3}-B=\dfrac{4}{3}-\dfrac{1}{x-\sqrt[]x+1}$
$=\dfrac{4(x-\sqrt[]x+1)-3}{3.(x-\sqrt[]x+1}$
$=\dfrac{4x-4\sqrt[]x+1}{3.(x-\sqrt[]x+1}$
$=\dfrac{(2\sqrt[]x-1)^2}{3.(x-\sqrt[]x+1}$
Mà $(2\sqrt[]x-1)^2≥0∀x$
$x-\sqrt[]x+1=(\sqrt[]x-\dfrac{1}{2})^2+\dfrac{3}{4}>0∀x$
$⇒\dfrac{(2\sqrt[]x-1)^2}{3.(x-\sqrt[]x+1}≥0∀x$
Hay $\dfrac{4}{3}-B≥0$
$⇒B≤\dfrac{4}{3}$
Dấu $=$ xảy ra $⇔2.\sqrt[]x-1=0$
$⇔\sqrt[]x=\dfrac{1}{2}$
$⇒x=\dfrac{1}{4}$
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