3 - ( x² + 2x )² + 2x² + 4x  \(\ge\) 0    \(\Leftrightarrow\left(x^2+2x\right)^2+2\left(x^2+2x\right)-3\le0.\)  Đặt  t = x² + 2x  = (x + 1)² - 1 ,      \(t\ge-1.\) 

BPT trở thành : \(\hept{\begin{cases}t^2+2t-3\le0\\t=(x+1)^2-1\ge-1\end{cases}\Leftrightarrow\hept{\begin{cases}-3\le t\le1\\t\ge-1\end{cases}\Leftrightarrow}-1\le t\le1.}\) 

Vậy ta có : \(-1\le x^2+2x\le1\Leftrightarrow x^2+2x-1\le0\Leftrightarrow-1-\sqrt{2}\le x\le-1+\sqrt{2}.\)

a. Ta có \(\left(x^2+1\right)\left(4x-2\right)\ge0\)

Mà \(x^2+1\ge0+1>0\)

\(\Leftrightarrow4x-2\ge0\Leftrightarrow x\ge\frac{1}{2}\)

b.Ta có: \(\left(x-2\right)x^2>0\)

mà \(x^2\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x^2\ne0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x>2\end{cases}\Leftrightarrow}x>2}\)

a: =>(x-1)(x-2)=0

=>x=1 hoặc x=2

b: TH1: x>=0

=>2x=3x+2

=>x=-2(loại)

TH2: x<0

=>-2x=3x+2

=>-5x=2

=>x=-2/5(nhận)

c: TH1: x>=0

=>2x=3x+4

=>-x=4

=>x=-4(loại)

TH2: x<0

=>-2x=3x+4

=>-5x=4

=>x=-4/5(nhận)

\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)

\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)

Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)

\[\left| {2x - 3} \right| > x + 1\\ \Leftrightarrow \left| {2x - 3} \right| - x > 1\\ T{H_1}:2x - 3 \ge 0 \Rightarrow x \ge {3 \over 2}\\ 2x - 3 - x > 1\\ \Leftrightarrow x - 3 > 1\\ \Leftrightarrow x > 4\left( {TM} \right)\\ T{H_2}:2x - 3 < 0 \Rightarrow x < {3 \over 2}\\ - \left( {2x - 3} \right) - x > 1\\ \Leftrightarrow - 2x + 3 - x > 1\\ \Leftrightarrow - 3x > - 2\\ \Leftrightarrow x < {2 \over 3}\left( {TM} \right)\]

2(x+4)(x-3)=0

=> (x+4)(x-3)=0

TH1: x+4=0 => x=-4

TH2: x-3=0=> x=3

vậy pt có nghiệm là ; -4;3

b) (x-1)²(3x-1)=0

TH1: x-1=0 => x=1

TH2:3x-1=0=>3x=1=>x=1/3

vậy pt có nghiệm là: 1;1/3

c) (2x/3 + 4)(2x-3) (x/2-1)=0

=> TH1:  2x/3  +4=0 => 2x/3 =-4 => 2x=-12 => x=-6

TH2: 2x-3=0 => 2x=3=>x=3/2

TH3:x/2 -1 =0 => x/2=1 => x=2

vậy pt có nghiệm là : -6;3/2;2

a, 2(x+4)(x-3)=0

 (x+4)(x+3)=0

x+4=0 hoặc x+3=0

x=-4 hoặc x=-3

b,(x-1)^2(3x-1)=0

x-1=0 hoặc 3x-1=0

x=1 hoặc x=1/3

c,(2x/3+4)(2x-3)(x/2-1)=0

2x/3+4=0 hoặc 2x-3=0 hoặc x/2-1=0

x=6 hoặc x=3/2 hoặc x=2

ĐKXĐ: \(x\ge3\)

\(\sqrt{x-1}>\sqrt{x-2}+\sqrt{x-3}\)

\(\Leftrightarrow x-1>2x-5+2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow4-x>2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\\left(4-x\right)^2>4\left(x^2-5x+6\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\3x^2-12x+8< 0\end{matrix}\right.\)

\(\Rightarrow\dfrac{6-2\sqrt{3}}{3}< x< \dfrac{6+2\sqrt{3}}{3}\)

Kết hợp ĐKXĐ \(\Rightarrow3\le x< \dfrac{6+2\sqrt{3}}{3}\)