a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)

a, (–31) . (x +7)=0

<=>  x +7 = 0

<=> x = -7

Vậy x \(\in\left\{-7\right\}\)

b, (8 – x) . (x + 13) = 0

<=> \(\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\)

Vậy x \(\in\left\{8;-13\right\}\)

c,(x²– 25) . (3– x )=0

<=> (x - 5) (x + 5) (3 - x) = 0

<=> \(\left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\)

Vậy x \(\in\left\{5;-5;3\right\}\)

d, ( x - 3 ) (x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=3\\x^2=-4\end{matrix}\right.\)_{(vô lý)}

Vậy x \(\in\left\{3\right\}\)

\(a,\left(x-8\right).\left(x+30\right)>0\Leftrightarrow\) \(\begin{cases}x-8>0\\x+30>0\end{cases}\)   \(\Rightarrow\)  \(\begin{cases}x>8\\x>-30\end{cases}\)  \(\Rightarrow x>8\)

                                   hoặc  \(\begin{cases}x-8< 0\\x+30< 0\end{cases}\)  \(\Rightarrow\)  \(\begin{cases}x< 8\\x< -30\end{cases}\)   \(\Rightarrow x< -30\)

Vậy \(\left(x-8\right).\left(x+30\right)>0\Leftrightarrow\)   x > 8 hoặc x < -30

\(b,\left(15-x\right)\left(x-3\right)>0\Leftrightarrow\)  \(\begin{cases}15-x>0\\x-3>0\end{cases}\)   \(\Rightarrow\)   \(\begin{cases}x< 15\\x>3\end{cases}\)   (TM)

                                 hoặc  \(\begin{cases}15-x< 0\\x-3< 0\end{cases}\)   \(\Rightarrow\)    \(\begin{cases}x>15\\x< 3\end{cases}\)  (loại)

Vậy \(\left(15-x\right)\left(x-3\right)>0\)  <=> x \(\in\)  {4;5;6...;14}

Ta có: 2 số cùng dấu luôn nhân lại đc 1 số lớn hơn 0.

a, ( x - 8 ) . ( x + 30 ) > 0

\(\Rightarrow\left[\begin{array}{nghiempt}\text{2 số cùng âm}\\\text{2 số cùng dương}\end{array}\right.\)

TH1 : \(\begin{cases}x-8< 0\\x+30< 0\end{cases}\)\(\Rightarrow\begin{cases}x< 8\\x< -30\end{cases}\)\(\Rightarrow x< -30\)

TH2: \(\begin{cases}x-8>0\\x+30>0\end{cases}\)\(\Rightarrow\begin{cases}x>8\\x>-30\end{cases}\)\(\Rightarrow x>8\)

Vậy để \(\text{( x - 8 ) . ( x + 30 ) > 0}\) thì \(\left[\begin{array}{nghiempt}x< -30\\x>8\end{array}\right.\)

b, ( 15 - x ) . ( x - 3 ) > 0

a: -26-(x-7)=0

=>x-7=-26

=>x=-26+7=-19

b: \(24-\left(30+x\right)=-7\)

=>\(30+x=24-\left(-7\right)\)

=>\(x+30=24+7=31\)

=>x=31-30=1

c: \(3-\left(17-x\right)=-12\)

=>\(17-x=3-\left(-12\right)\)

=>\(17-x=3+12=15\)

=>x=17-15=2

d: \(\left(5-x\right)-7=-15\)

=>\(5-x=-15+7=-8\)

=>\(x=5-\left(-8\right)=5+8=13\)

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

a) \(\left(x-17\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-15\end{matrix}\right.\)

b) \(\left(6-x\right)\left(x-35\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=35\end{matrix}\right.\)

a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)

\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)

\(\Leftrightarrow12x=-2\)

hay \(x=-\dfrac{1}{6}\)

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)

\(\Leftrightarrow2x=-40\)

hay x=-20

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