Lời giải đề 16-trang 1

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: $A=dfrac{sqrt{x}+1}{x+4sqrt{x}+4}:left$dfracxx+2sqrtx+dfracxsqrtx+2right$$$=dfrac{sqrt{x}+1}{{{$sqrtx+2$}^{2}}}:left$dfracxsqrtx(sqrtx+2)+dfracxsqrtx+2right$$

$=dfrac{sqrt{x}+1}{{{$sqrtx+2$}^{2}}}:left$dfracsqrtxsqrtx+2+dfracxsqrtx+2right$$$=dfrac{sqrt{x}+1}{{{$sqrtx+2$}^{2}}}:dfrac{sqrt{x}$sqrtx+1$}{sqrt{x}+2}$$=dfrac{1}{sqrt{x}$sqrtx+2$}$

Câu II.2:  Với $x>0$ ta có $A=dfrac{1}{sqrt{x}$sqrtx+2$}$ và $sqrt{x}>0$; $sqrt{x}+2>0$.

            Khi đó $Age frac{1}{3sqrt{x}}Leftrightarrow dfrac{1}{sqrt{x}left$sqrtx+2right$}ge dfrac{1}{3sqrt{x}}$ $Leftrightarrow sqrt{x}+2le 3$$Leftrightarrow sqrt{x}le 1$$Leftrightarrow xle 1$

Suy ra: $0<xle 1$.Câu III.1:  Do $left$dright$//left$d^{\prime }right$$ nên(left{ begin{align}   & a=2 \  & bne 3 \ end{align} right.).  Do $left(dright$)đi qua điểm $Aleft(1;-1right$) nên:

$-1=2.1+bLeftrightarrow b=-3$ $thỏamãnđiềukiện\$bne3\$$ .

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 $

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 $m=2$.