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+4sqrt{x}+4}:left( dfrac{x}{x+2sqrt{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}$$=dfrac{1}{sqrt{x}(sqrt{x}+2)}$
Câu II.2: Với $x>0$ ta có $A=dfrac{1}{sqrt{x}(sqrt{x}+2)}$ và $sqrt{x}>0$; $sqrt{x}+2>0$.
Khi đó $Age frac{1}{3sqrt{x}}Leftrightarrow dfrac{1}{sqrt{x}left( sqrt{x}+2 right)}ge dfrac{1}{3sqrt{x}}$ $Leftrightarrow sqrt{x}+2le 3$$Leftrightarrow sqrt{x}le 1$$Leftrightarrow xle 1$
Suy ra: $0<xle 1$.Câu III.1: Do $left( d right)//left( d’ right)$ nên(left{ begin{align} & a=2 \ & bne 3 \ end{align} right.). Do (left( d right))đi qua điểm (Aleft( 1;-1 right)) nên:
$-1=2.1+bLeftrightarrow b=-3$ (thỏa mãn điều kiện $bne 3$) .
Vậy $a=2$, $b=-3$.
Câu III.2: Điều kiện có nghiệm: $Delta ={{(m-2)}^{2}}+12>0,,,forall m$
Ta có: $sqrt{x_{1}^{2}+2018}-{{x}_{1}}=sqrt{x_{2}^{2}+2018}+{{x}_{2}}$ $Leftrightarrow sqrt{x_{1}^{2}+2018}-sqrt{x_{2}^{2}+2018}={{x}_{2}}+{{x}_{1}}$
$Leftrightarrow dfrac{x_{1}^{2}-x_{2}^{2}}{sqrt{x_{1}^{2}+2018}+sqrt{x_{2}^{2}+2018}}={{x}_{2}}+{{x}_{1}}$
$Leftrightarrow $ (left[ begin{align} & {{x}_{1}}+{{x}_{2}}=0text{ (1)} \ & sqrt{x_{1}^{2}+2018}+sqrt{x_{2}^{2}+2018}={{x}_{1}}-{{x}_{2}}text{ },text{(2)} \ end{align} right.)
Theo định lí Viet ta có: ${{x}_{1}}+{{x}_{2}}=m-2$. Khi đó: (1) $Leftrightarrow $$m-2=0Leftrightarrow 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 $m=2$.
