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Đặt: $\left\{\begin{array}{l}x=2a\\y=3b\\z=2a+3c\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\dfrac{x}{2}\\b=\dfrac{y}{3}\\c=\dfrac{z-x}{3}\end{array}\right.$
Khi đó, ta có:
$P=\dfrac{y+z-x}{x}+\dfrac{2x+z-x}{y}+\dfrac{4(y+x-z)}{z}$
     $=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{4x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{4y}{z}\right)-5$
     $\ge 2\sqrt{\dfrac{y}{x}.\dfrac{x}{y}}+2\sqrt{\dfrac{z}{x}.\dfrac{4x}{z}}+2\sqrt{\dfrac{z}{y}.\dfrac{4y}{z}}-5=5$
Dấu bằng xảy ra khi: $2x=2y=z \Leftrightarrow \dfrac{a}{3}=\dfrac{b}{2}=\dfrac{c}{2}$

b.
$\mathop {\lim }\limits_{x \to 3^+}\dfrac{10+3|2-x^2|}{4x-|2x^2-3x+1|}=\dfrac{10+3|2-3^2|}{4.3-|2.3^2-3.3+1|}=\dfrac{31}{2}$

a.
$\mathop {\lim }\limits_{x \to 2^+}\dfrac{10-3|x-3|}{5x-4}=\dfrac{10-3|2-3|}{5.2-4}=\dfrac{7}{2}$

a.
Ta có:
      $\left\{\begin{array}{l}x^3+y^3=1\\x^2y+2xy^2+y^3=2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3=1\\x^2y+2xy^2+y^3=2(x^3+y^3)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3\\2x^3-x^2y-2xy^2+y^3=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^3+y^3=1\\(x-y)(x+y)(2x-y)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x^3+y^3=1\\x=y\end{array}\right.\\\left\{\begin{array}{l}x^3+y^3=1\\2x=y\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=\dfrac{1}{\sqrt[3]2}\\y=\dfrac{1}{\sqrt[3]2}\end{array}\right.\\\left\{\begin{array}{l}x=\dfrac{2}{\sqrt[3]9}\\y=\dfrac{2}{\sqrt[3]9}\end{array}\right.\end{array}\right.$

a.
Điều kiện: $x,y\ge0$.
Hệ đã cho tương đương với:
      $\left\{\begin{array}{l}x\sqrt x-y\sqrt y=\sqrt x+8\sqrt y\\x-y=5\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}5(x\sqrt x-y\sqrt y)=(\sqrt x+8\sqrt y)(x-y)\\x-y=5\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}4x\sqrt x-8x\sqrt y+y\sqrt x+3y\sqrt y=0\\x-y=5\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(\sqrt x-\sqrt y)(2\sqrt x-3\sqrt y)(2\sqrt x+\sqrt y)=0\\x-y=5\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}4x=9y\\x-y=5\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x=9\\y=4\end{array}\right.$

2.
Ta có:
$\dfrac{x-1}{x-2}\le0 \Leftrightarrow 1\le x<2$
$4x+1\le m \Leftrightarrow x\le\dfrac{m-1}{4}$

a. Hệ có nghiệm
$\Leftrightarrow \dfrac{m-1}{4}\ge1$
$\Leftrightarrow m-1\ge4$
$\Leftrightarrow m\ge5$

b. Hệ có nghiệm duy nhất
$\Leftrightarrow \dfrac{m-1}{4}=1$
$\Leftrightarrow m=5$