\(\left(x+3\right)\left(2x-4\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x+3>0\\2x-4< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\) \(\Leftrightarrow-3< x< 2\)

TH2: \(\left\{{}\begin{matrix}x+3< 0\\2x-4>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>2\end{matrix}\right.\)\(\Leftrightarrow x\in\varnothing\)

Vậy \(-3< x< 2\)

Có: (x + 3)(2x - 4) < 0 \(\Leftrightarrow\) x + 3 và 2x - 4 trái dấu

+) Xét: x + 3 < 0 => x < -3

2x - 4 > 0=>  x > 2 

=> 2 < x < -3 (vô lí) => loại

+) Xét: x + 3 > 0 => x > -3  

2x - 4 < 0 => x < 2

=> -3 < x < 2 (tm)

Vậy... 

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

\(\left(x+\frac{2}{3}\right)\left(\frac{5}{4}-2x\right)>0\)

th1 : 

\(\hept{\begin{cases}x+\frac{2}{3}>0\\\frac{5}{4}-2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{2}{3}\\-2x>-\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x>-\frac{2}{3}\\x>\frac{5}{8}\end{cases}\Rightarrow}x>\frac{5}{8}}\)

th2 : 

\(\hept{\begin{cases}x+\frac{2}{3}< 0\\\frac{5}{4}-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{2}{3}\\-2x< -\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x< -\frac{2}{3}\\x< \frac{5}{8}\end{cases}\Rightarrow}x< -\frac{2}{3}}\)

vậy_

(2x+4).(x+4)<0

=> Phải có 1 thừa số lớn hơn 0 và 1 thừa số nhỏ hơn 0

TH1

2x+4>0 và x+4<0

2x+4>0 => 2x>-4  =>x>-2 (1)

x+4<0 =>x<-4 (2)

(1) và (2) ko thể cùng xảy ra, vậy TH1 loại

TH2 2x+4<0 và x+4>0

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

x+4>0 =>x>-4  (4)

Từ (3) và (4) => -2<x<-4

Vậy x=-3

\(\left(2x+4\right)\left(x+4\right)< 0\)

\(\Rightarrow2\left(x+2\right)\left(x+4\right)< 0\)

\(\Rightarrow\left(x+2\right)\left(x+4\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x+2< 0\\x+4>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+2>0\\x+4< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< -2\\x>-4\end{cases}}\) hoặc \(\hept{\begin{cases}x>-2\\x< -4\end{cases}}\left(loại\right)\)

\(\Rightarrow-4< x< -2\)

Mà \(x\in Z\)

\(\Rightarrow x=-3\)

x=-3 nha bạn 

chúc bạn học giỏi nha bạn

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

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

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x-6=0\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}\)

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2\right)+\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-2+x+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

(2x-3)(x+1)+(2x+3)(3x-7)=0

<=> (2x-3)(x+1)-(2x-3)(3x-7)=0

<=> (2x-3)(x+1-3x+7)=0

<=> (2x-3)(-2x+8)=0

<=> 2x-3=0 => x=3/2

Hoặc -2x+8=0 => x= 4

Vậy x thuộc{3/2;4}

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}\)

(2x – 1)2 – (x + 3)2 = 0

[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0 (phân tích ra hằng đẳng thức số 3)

(2x - 1 - x - 3)(2x - 1 + x + 3) = 0 (bỏ ngoặc)

(x - 4)(3x + 2) = 0 (rút gọn)

Hoặc x - 4 = 0 => x = 4

=> A=0 hoặc B=0 Hoặc 3x + 2 = 0 => 3x = 2 => x = -2/3

 Vậy x = { 4 ; -2/3} 

Đề:  (2x-1)^2 - (x+3)^2=0

3x^2= 10x + 8

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

3 x^ 2 - 10x - 8 =0

=> x= âm 2/3

x = 4

(2x-1)^2=0 hoặc (x+3)^2=0

2x-1=0             x+3=0

2x=0+1           x=-3

x=1/2