\(S=\dfrac{x}{2}+\dfrac{1}{2x}+\dfrac{y}{2}+\dfrac{2}{y}+\dfrac{1}{2}\left(x+y\right)\)

\(S\ge2\sqrt{\dfrac{x}{4x}}+2\sqrt{\dfrac{2y}{2y}}+\dfrac{1}{2}.3=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le m\end{matrix}\right.\)

Hệ có nghiệm duy nhất \(\Leftrightarrow m=2\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x\ge6\\2x\le6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{6}{m^2+1}\\x\le3\end{matrix}\right.\)

Hệ có nghiệm duy nhất \(\Leftrightarrow\dfrac{6}{m^2+1}=3\)

\(\Leftrightarrow m=\pm1\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-6x+9\ge x^2+7x+1\\5x\ge2m-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{8}{13}\\x\ge\dfrac{2m-8}{5}\end{matrix}\right.\)

Pt có nghiệm duy nhất khi \(\dfrac{2m-8}{5}=\dfrac{8}{13}\Leftrightarrow m=\dfrac{72}{13}\)

d.

Hệ có nghiệm duy nhất khi:

TH1:

 \(\left\{{}\begin{matrix}m>0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-9=m^2-9m\end{matrix}\right.\) \(\Leftrightarrow m=1\)

TH2:

\(\left\{{}\begin{matrix}m+3< 0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)

\(\Leftrightarrow m=1\) (ktm)

Vậy \(m=1\)

e.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m-1\right)x\ge-2m+3\\\left(4-4m\right)x\le3\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi:

\(\left\{{}\begin{matrix}\left(2m-1\right)\left(4-4m\right)>0\\\dfrac{-2m+3}{2m-1}=\dfrac{3}{4-4m}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}< m< 1\\\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow m=\dfrac{3}{4}\)

Mình nghĩ là thế này

Ta có: x²+1>0 ∀xϵR

x²+2x+3=(x+1)²+1>0 ∀xϵR

x²+4x+5=(x+2)²+1 >0 ∀xϵR

nên \(\sqrt{x^2+1}+2\sqrt{x^2+2x+3}\ge3\sqrt{x^2+4x+5}\)

\(\Leftrightarrow\sqrt{x^2+1}+2\sqrt{\left(x+1\right)^2+1}\ge3\sqrt{\left(x+2\right)^2+1}\)

\(\Leftrightarrow x+1+2\left(x+1\right)+2\ge3\left(x+2\right)+3\)

\(\Leftrightarrow x+3+2x+2\ge3x+6+3\)

\(\Leftrightarrow3x+5\ge3x+9\Leftrightarrow0x\ge4\) (vô nghiệm)

Vậy S=∅

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x^2+2x+3}=b>0\end{matrix}\right.\)

\(a+2b\ge3\sqrt{2b^2-a^2}\)

\(\Leftrightarrow a^2+4b^2+4ab\ge18b^2-9a^2\)

\(\Leftrightarrow5a^2+2ab-7b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(5a+7b\right)\ge0\)

\(\Leftrightarrow a-b\ge0\) (do \(5a+7b>0\))

\(\Leftrightarrow a\ge b\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x^2+2x+3}\)

\(\Leftrightarrow x^2+1\ge x^2+2x+3\Leftrightarrow x\le-1\)

Vậy nghiệm của BPT là \(x\le-1\)

\(\left\{{}\begin{matrix}2x-1\ge3x-9\\2-x< 2x-6\\x-3\ge4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3x\ge-9+1\\-x-2x< -6-2\\x\ge4+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\ge-8\\-3x< -8\\x\ge7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x>\dfrac{8}{3}\\x\ge7\end{matrix}\right.\Leftrightarrow7\le x\le8\)

BĐT\(\Leftrightarrow\left(\frac{1}{x-1}\right)^3+\left(\frac{x-1}{y}\right)^3+\left(\frac{1}{y}\right)^3\ge3\left(\frac{1}{x-1}+\frac{x-1}{y}+\frac{1}{y}-2\right)\)

Đặt \(\left(\frac{1}{x-1};\frac{x-1}{y};\frac{1}{y}\right)=\left(a;b;c\right)\)

BĐT cần cm \(\Leftrightarrow a^3+b^3+c^3\ge3\left(a+b+c-2\right)\)

\(\Leftrightarrow\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3\left(a+b+c\right)\)

Đúng theo AM-GM --> đpcm

Chọn C

Lời giải:

$2x+5< x+8-m$

$\Leftrightarrow 2x-x< 8-5-m$

$\Leftrightarrow x< 3-m$

Để BPT có nghiệm đều là số âm thì $3-m\leq 0$

$\Lefrightarrow m\geq 3$ 

Đáp án D.