Câu I.1: PT ${{x}^{2}}+8x+7=0$ có $a-b+c=1-8+7=0$nên có hai nghiệm $x=-1$; $x=-7$.
Câu I.2: \(\left\{ \begin{align} & 2x-y=-6 \\ & 5x+y=20 \\ \end{align} \right.\Leftrightarrow \)\(\left\{ \begin{align} & 7x=14 \\ & 5x+y=20 \\ \end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align} & x=2 \\ & 5x+y=20 \\ \end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align} & x=2 \\ & 10+y=20 \\ \end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align} & x=2 \\ & y=10 \\ \end{align} \right.\)
Câu II.1: Ta có: $A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left( \dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2} \right)$$=\dfrac{\sqrt{x}+1}{{{(\sqrt{x}+2)}^{2}}}:\left( \dfrac{x}{\sqrt{x}(\sqrt{x}+2)}+\dfrac{x}{\sqrt{x}+2} \right)$
$=\dfrac{\sqrt{x}+1}{{{(\sqrt{x}+2)}^{2}}}:\left( \dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{x}{\sqrt{x}+2} \right)$$=\dfrac{\sqrt{x}+1}{{{(\sqrt{x}+2)}^{2}}}:\dfrac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}+2}$$=\dfrac{1}{\sqrt{x}(\sqrt{x}+2)}$
Câu II.2: Với $x>0$ ta có $A=\dfrac{1}{\sqrt{x}(\sqrt{x}+2)}$ và $\sqrt{x}>0$; $\sqrt{x}+2>0$.
Khi đó $A\ge \frac{1}{3\sqrt{x}}\Leftrightarrow \dfrac{1}{\sqrt{x}\left( \sqrt{x}+2 \right)}\ge \dfrac{1}{3\sqrt{x}}$ $\Leftrightarrow \sqrt{x}+2\le 3$$\Leftrightarrow \sqrt{x}\le 1$$\Leftrightarrow x\le 1$
Suy ra: $0<x\le 1$.Câu III.1: Do $\left( d \right)//\left( d' \right)$ nên\(\left\{ \begin{align} & a=2 \\ & b\ne 3 \\ \end{align} \right.\). Do \(\left( d \right)\)đi qua điểm \(A\left( 1;-1 \right)\) nên:
$-1=2.1+b\Leftrightarrow b=-3$ (thỏa mãn điều kiện $b\ne 3$) .
Vậy $a=2$, $b=-3$.
Câu III.2: Điều kiện có nghiệm: $\Delta ={{(m-2)}^{2}}+12>0,\,\,\forall m$
Ta có: $\sqrt{x_{1}^{2}+2018}-{{x}_{1}}=\sqrt{x_{2}^{2}+2018}+{{x}_{2}}$ $\Leftrightarrow \sqrt{x_{1}^{2}+2018}-\sqrt{x_{2}^{2}+2018}={{x}_{2}}+{{x}_{1}}$
$\Leftrightarrow \dfrac{x_{1}^{2}-x_{2}^{2}}{\sqrt{x_{1}^{2}+2018}+\sqrt{x_{2}^{2}+2018}}={{x}_{2}}+{{x}_{1}}$
$\Leftrightarrow $ \(\left[ \begin{align} & {{x}_{1}}+{{x}_{2}}=0\text{ (1)} \\ & \sqrt{x_{1}^{2}+2018}+\sqrt{x_{2}^{2}+2018}={{x}_{1}}-{{x}_{2}}\text{ }\,\text{(2)} \\ \end{align} \right.\)
Theo định lí Viet ta có: ${{x}_{1}}+{{x}_{2}}=m-2$. Khi đó: (1) $\Leftrightarrow $$m-2=0\Leftrightarrow m=2$.
Do $\sqrt{x_{1}^{2}+2018}>\left| {{x}_{1}} \right|$; $\sqrt{x_{2}^{2}+2018}>\left| {{x}_{2}} \right|$ suy ra $\sqrt{x_{1}^{2}+2018}+\sqrt{x_{2}^{2}+2018}>\left| {{x}_{1}} \right|+\left| {{x}_{2}} \right|\ge {{x}_{1}}-{{x}_{2}}$ nên (2) không xảy ra.
Vậy $m=2$.
