a, (3x-2)(4x+5)=0

↔ TH1: 3x-2 = 0 ↔ x = 2/3

  TH2 : 4x+5 = 0 ↔ x = -5/4

Vậy PT có tập no S = ( 2/3; -5/4)

b,(2,3x-6,9)(0,1x+2)=0

↔ TH1: 2,3x - 6,9 = 0 ↔ x = 3

    TH2 : 0,1x + 2 = 0 ↔ x = -20

Vậy PT có tập no S = ( 3; -20)

c, (4x+2)(x^2 +1)=0

TH1: 4x+2=0 ↔ x = -1/2

Th2 : x^2 +1≠0 ( vô lí)

Vậy PT có tập no S = (-1/2)

d, (2x+7)(x-5)(5x+1)=0

↔ TH1: 2x+7 = 0 ↔ x = -7/2

 TH2: x-5 = 0 ↔ x = 5

TH3 :  5x+1 = 0 ↔ x = -1/5

Vậy PT có tập no S = ( -7/2 ; 5 ; -1/5

a, \(\left(3x-2\right)\left(4x+5\right)=0\Leftrightarrow x=\frac{2}{3};x=-\frac{5}{4}\)

b, \(\left(2,3-6,9\right)\left(0,1x+2\right)=0\Leftrightarrow\frac{x}{10}+2=0\Rightarrow x=-20\)

c, \(\left(4x+2\right)\left(x^2+1>0\right)=0\Leftrightarrow x=-\frac{1}{2}\)

a)

\(x^2-5x+4x-20=0.\)

\(x^2-x-20=0\)

\(\left(x^2-x+\frac{1}{4}\right)-20-\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2-\left(\frac{20.4+1}{4}\right)=0\)

\(\hept{\begin{cases}x-\frac{1}{2}-\left(\frac{20.4+1}{4}\right)=0\\x-\frac{1}{2}+\left(\frac{20.4+1}{4}\right)=0\end{cases}}\)

b)  \(x^2+6x-7x-42=0\)

\(x^2-x-42=0\)

\(x^2-x+\frac{1}{4}-42-\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2-\left(\frac{42.4+1}{4}\right)=0\)  " tương tự con A

\(x^3-16x=0\)

\(x\left(x^2-16\right)=0\)

\(x=0,+4,-4\)

\(x^3-16x=0\)

\(x.\left(x^2-16\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=\pm4\)

Tham khảo nhé~

\(4x^2-28=0\)

\(4x^2=28\)

\(x^2=7\)

\(\)

\(4x^2-28=0\)

\(\Leftrightarrow4\left(x^2-7\right)=0\)

\(\Leftrightarrow x^2-7=0\)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow x=\pm\sqrt{7}\)

\(x^2+36=12x\)

\(\Rightarrow x^2+36-12x=0\)

\(\Rightarrow\left(x-6\right)^2=0\)

\(\Rightarrow x-6=0\)

\(\Rightarrow x=6\)

\(\left(x-7\right)\left(x+2019\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+2019=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2019\end{cases}}\)

\(9-25=\left(7-x\right)-\left(25+7\right)\)

\(\Leftrightarrow-16=7-x-25-7\)

\(\Leftrightarrow-x=-16+25\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

\(2\left(4x-2x\right)-7x=15\)

\(\Leftrightarrow4x-7x=15\)

\(\Leftrightarrow x=-5\)

a ) 9 - 25 = ( 7 - x ) - ( 25 + 7 )

    9 - 25  =  7 - x - 25 - 7 

  9 - 25 - 7 + 25 + 7 = -x 

9        = - x

 => x = -9

Vậy x = -9

b) 2 . ( 4x - 2x ) - 7x = 15

    8x  - 4x - 7x          = 15 

-3x = 15

  x   =  15 : ( - 3 ) 

  x = -5 

Vậy x = -5

c ) ( x - 7 ). ( x + 2019 ) = 0

 => x - 7 = 0 hoặc x + 2019 = 0

   => x    = 7 hoặc x = - 2019 

 vậy x \(\in\){ 7 ; -2019 }

a. 9 - 25 = (7 - x) - (25 + 7)

=> -16 = 7 - x - 25 - 7

=> -16 = -x - 25

=> -x = 9

=> x = -9

b. 2 . (4x - 2x) - 7x = 15

=> 2 . 2x - 7x = 15

=> 4x - 7x = 15

=> -3x = 15

=> x = -5

c. (x - 7) . (x + 2019) = 0

=> x - 7 = 0 hoặc x + 2019 = 0

=> x = 7 hoặc x = -2019

a) Ta có :  \(x^2-6x+10\)

\(=\left(x^2-6x+9\right)+1\)

\(=\left(x-3\right)^2+1\ge1>0\forall x\)

b) Ta có :  \(4x-x^2-5\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)

Vậy ...

Để ;(x + 1).(x - 3) < 0 thì ta có 2 trường hợp

Th1 : \(\hept{\begin{cases}x+1< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>3\end{cases}\left(loai\right)}}\)

Th2 : \(\hept{\begin{cases}x+1>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow}-1< x< 3}\)

Ta có: \(x^2-4x+5=\left(x^2-2.x.2+2^2\right)+1\)

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

Vì \(\left(x-2\right)^2\ge0\left(\forall x\right);1\ge0\)

Vậy \(x^2-4x+5\ge0\left(\forall x\right)\)