\(a,\left|3x-1\right|=\left|5-2x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=5-2x\\3x-1=2x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=6\\x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{6}{5}\\x=-4\end{cases}}\)

b,\(\left|2x-1\right|+x=2\)

\(\Leftrightarrow\left|2x-1\right|=2-x\)

Điều kiện \(2-x\ge0\Leftrightarrow x\le2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=2-x\\2x-1=x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=3\\x=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\left(\text{nhận}\right)\\x=-1\left(\text{nhận}\right)\end{cases}}}\)

c.\(A=0,75-\left|x-3,2\right|\)

Vì \(\left|x-3,2\right|\ge0\Rightarrow0,75-\left|x-3,2\right|\le0,75\)

Dấu "=' xảy ra \(\Leftrightarrow x-3,2=0\Leftrightarrow x=3,2\)

Vậy Max A = 0,75 khi x = 3,2

\(d,B=2.\left|x+1,5\right|-3,2\)

Vì 2. |x + 1,5| ≥ 0 => B ≥ -3,2

Dấu " = ' xảy ra khi \(2\left|x+1,5\right|=0\)

\(\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\)

Vậy Min B = -3,2 khi x = -1,5

\(b,B\left(x\right)=x\left(x-3\right)-2\left(x+5\right)=x^2-3x-2x-10=x^2-5x-10\)

\(=x^2-\frac{5}{2}x-\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-10=x\left(x-\frac{5}{2}\right)-\frac{5}{2}\left(x-\frac{5}{2}\right)-\frac{65}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{65}{4}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0=>\left(x-\frac{5}{2}\right)^2-\frac{65}{4}\ge-\frac{65}{4}\) (với mọi x)

Dấu "=" xảy ra \(< =>x-\frac{5}{2}=0< =>x=\frac{5}{2}\)

Vậy minB(x)=-65/4 khi x=5/2

\(c,C\left(x\right)=2x\left(x+1\right)-3x\left(x+1\right)=2x^2+2x-3x^2-3x=-x^2-x\)

\(=-\left(x^2+x\right)=-\left(x^2+x+1-1\right)=-\left(x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}-1\right)\)

\(=-\left[x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)-\frac{1}{4}\right]=-\left[\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right]=\frac{1}{4}-\left(x+\frac{1}{2}\right)^2\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0=>\frac{1}{4}-\left(x+\frac{1}{2}\right)^2\le\frac{1}{4}\) (với mọi x)

Dấu  "=" xảy ra \(< =>x+\frac{1}{2}=0< =>x=-\frac{1}{2}\)

Vậy maxC(x)=1/4 khi x=-1/2

\(A\left(x\right)=2x\left(x-1\right)-3\left(x-13\right)=2x^2-5x+39\)

\(=2\left(x^2-\frac{5}{2}x+\frac{39}{2}\right)=2\left(x^2-\frac{5}{4}x-\frac{5}{4}x+\frac{25}{16}-\frac{25}{16}+\frac{39}{2}\right)\)

\(=2\left[x\left(x-\frac{5}{4}\right)-\frac{5}{4}\left(x-\frac{5}{4}\right)\right]+\frac{287}{16}=2\left[\left(x-\frac{5}{4}\right)^2+\frac{287}{16}\right]=2\left(x-\frac{5}{4}\right)^2+\frac{287}{8}\)

Vì \(2\left(x-\frac{5}{4}\right)^2\ge0=>2\left(x-\frac{5}{4}\right)^2+\frac{287}{8}\ge\frac{287}{8}>0\) với mọi x

=>A(x) vô nghiệm (đpcm)

Ta có: |x+1,5|>=0(với mọi x)

=> 2|x+1,5|>=0

=> 2|x+1,5|-3,2>=-3,2 hay B>=-3,2

Do đó, GTNN của B là -3,2 khi: |x+1,5|=0

x+1,5=0

x=0-1,5

x=-1,5

Vậy GTNN của B là -3,2 khi x=-1,5

\(a,\Leftrightarrow-4+k=-3\Leftrightarrow k=1\\ b,\Leftrightarrow-3\left(2k-18\right)=40\\ \Leftrightarrow2k-18=-\dfrac{40}{3}\Leftrightarrow k=\dfrac{7}{3}\\ c,\Leftrightarrow10+18=9\left(2+k\right)\\ \Leftrightarrow k+2=\dfrac{28}{9}\Leftrightarrow k=\dfrac{10}{9}\)