TH1: x>=-5

=>3x+1=x+5

=>2x=4

=>x=2(nhận)

TH2: x<-5

=>-3x-1=x+5

=>-4x=6

=>x=-3/2(loại)

a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

\(Q=\dfrac{x\left(-x^2-2x-6\right)}{x}\)

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

\(Q=-\left(x^2+2x+6\right)\)

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)

a: Ta có: \(5-3x< 8\)

\(\Leftrightarrow3x>-3\)

hay x>-1

b: Ta có: \(\dfrac{2x-5}{4}\ge\dfrac{3-x}{3}\)

\(\Leftrightarrow3\left(2x-5\right)\ge4\left(3-x\right)\)

\(\Leftrightarrow6x-15\ge12-4x\)

\(\Leftrightarrow10x\ge27\)

hay \(x\ge\dfrac{27}{10}\)

c: Ta có: \(2x+5< x+7\)

\(\Leftrightarrow2x-x< 7-5\)

hay x<2

d: Ta có: \(4\left(x-3\right)\ge x+2\)

\(\Leftrightarrow4x-12-x-2\ge0\)

\(\Leftrightarrow3x\ge14\)

hay \(x\ge\dfrac{14}{3}\)

e: Ta có: \(\dfrac{2x+2}{3}< 2+\dfrac{x-2}{2}\)

\(\Leftrightarrow4x+4< 12+3x-6\)

\(\Leftrightarrow4x-3x< 6-4\)

hay x<2

f: Ta có: \(x-\dfrac{5x+2}{6}>\dfrac{7-3x}{4}\)

\(\Leftrightarrow12x-2\left(5x+2\right)>3\left(7-3x\right)\)

\(\Leftrightarrow2x-4>21-9x\)

\(\Leftrightarrow11x>25\)

hay \(x>\dfrac{25}{11}\)

mình làm 1 câu mấy câu còn lại tương tự nhé

\(A=-\dfrac{1}{x^2+12x-3}\)Ta có \(x^2+12x-3=\left(x+6\right)^2-39\ge-39\Rightarrow A\ge\dfrac{1}{39}\)

Dấu ''='' xảy ra khi x = -6 

Vậy với x = -6 thì A đạt GTNN là 1/39 

a: x^2+4x-9=x^2+4x+4-13=(x+2)^2-13>=-13

=>B<=-1/13

Dấu = xảy ra khi x=-2

b: 9x^2-6x+12=(3x-1)^2+11>=11

=>C<=3/11

Dấu = xảy ra khi x=1/3

c: 4x^2-12x+5=(2x-3)^2-4>=-4

=>D<=-2021/4

Dấu = xảy ra khi x=3/2

a) Ta có: \(2x+x^2=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b) Ta có: \(\left(2x+1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)