\(\text{f(x)}\)\(\text{>0}\)\(\text{⇔}\)\(\text{2x}\)²\(\text{-3x+1}\)\(>0\)⇔\(\left\{{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\)

⇒x∈(−∞;\(\dfrac{1}{2}\))∪(1;+∞)

f(x)>0⇔4-2x>0⇔x<2⇒x∈(−∞;2) 

 \(VT=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)

\(VP=-4x^2+12x-9-1=-\left(2x-3\right)^2-1\le-1\)

\(\Rightarrow VT>VP\)  ; \(\forall x\)

\(\Rightarrow\) Pt đã cho luôn luôn vô nghiệm

b.

\(\Leftrightarrow\left(m^2+3m\right)x=-m^2+4m+21\)

\(\Leftrightarrow m\left(m+3\right)x=\left(7-m\right)\left(m+3\right)\)

Để pt có nghiệm duy nhất \(\Rightarrow m\left(m+3\right)\ne0\Rightarrow m\ne\left\{0;-3\right\}\)

Khi đó ta có: \(x=\dfrac{\left(7-m\right)\left(m+3\right)}{m\left(m+3\right)}=\dfrac{7-m}{m}\)

Để nghiệm pt dương

\(\Leftrightarrow\dfrac{7-m}{m}>0\Leftrightarrow0< m< 7\)

\(m^2\left(x-1\right)+x-3< 0\Leftrightarrow\left(m^2+1\right)x-m^2-3< 0\)

Đặt \(f\left(x\right)=\left(m^2+1\right)x-m^2-3\)

\(f\left(x\right)< 0\forall x\in\left[-5;2\right]\Leftrightarrow\hept{\begin{cases}f\left(-5\right)< 0\\f\left(2\right)< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}-6m^2-8< 0\\m^2-1< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6m^2+8>0\\m^2< 1\end{cases}}\Leftrightarrow\left|m\right|< 1\Leftrightarrow-1< m< 1\)

Vậy có duy nhất 1 giá trị nguyên của tham số m thỏa mãn yêu cầu bài toán, đó là giá trị m = 0

Lời giải:

\(f(x)=(-x+1)(x-2)>0\Leftrightarrow \left\{\begin{matrix} -x+1< 0\\ x-2< 0\end{matrix}\right.\) hay $1< x< 2$

hay $x\in (1;2)$

Đáp án D

\(f\left(x\right)=\left(3m-4\right)x^2-2\left(m-2\right)x+m-1< 0\)

\(TH1:3m-4=0\Leftrightarrow m=\dfrac{4}{3}\Rightarrow f\left(x\right)=\dfrac{4}{3}x+\dfrac{1}{3}< 0\Leftrightarrow x< -\dfrac{1}{4}\left(ktm\right)\)

\(TH2:3m-4>0\Leftrightarrow m>\dfrac{4}{3}\Rightarrow f\left(x\right)< 0\forall x>1\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\x1\le1< x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)^2-\left(m-1\right)\left(3m-4\right)>0\\\left(x1-1\right)\left(x2-1\right)\le0\Leftrightarrow x1.x2-\left(x1+x2\right)+1\le0\\\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0< m< \dfrac{3}{2}\\\dfrac{m-1}{3m-4}-\dfrac{2\left(m-2\right)}{3m-4}+1\le0\Leftrightarrow\dfrac{1}{2}\le m< \dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{2}\le m< \dfrac{4}{3}\left(màm>\dfrac{4}{3}\right)\Rightarrow loại\)

\(TH3:3m-4< 0\Leftrightarrow m< \dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\Delta'=0\Leftrightarrow m=0\left(tm\right)\\x=\dfrac{2\left(m-2\right)}{3m-4}=\dfrac{1}{2}\notin\left(1;+\infty\right)\left(tm\right)\end{matrix}\right.\\\Delta'< 0\Leftrightarrow\left[{}\begin{matrix}m< 0\\m>\dfrac{3}{2}\end{matrix}\right.\\x1< x2\le1\left(1\right)\\\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\Leftrightarrow0< m< \dfrac{3}{2}\\\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0< m< \dfrac{3}{2}\\\dfrac{m-1}{3m-4}-\dfrac{2\left(m-2\right)}{3m-2}+1\ge0\\\dfrac{2\left(m-2\right)}{3m-4}-2< 0\end{matrix}\right.\)

\(\Leftrightarrow0< m\le\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}m\le0\\0< m\le\dfrac{1}{2}\end{matrix}\right.\)

=>2x+6y=2m+2 và 2x-y=7

=>7y=2m-5 và 2x-y=7

=>y=2/7m-5/7 và 2x=y+7

=>y=2/7m-5/7 và 2x=2/7m+30/7

=>x=1/7m+15/7 và y=2/7m-5/7

x0+2y0 bằng gì bạn ơi?