a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

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

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay 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 \(\in\left\{3\right\}\)

Bài 1.

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

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

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

\(1,\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\left(B\right)\\ 2,Giống.1\\ 3,=5x^2\left(B\right)\\ 4,x^3=x\Leftrightarrow x^3-x=0\\ \Leftrightarrow x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow A\)

\(x^2-25+x\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)+x\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-5+x\right)=0\\ \Leftrightarrow\left(x+5\right)\left(2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{5}{2}\end{matrix}\right.\)

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

x.(x+5)-x\(^2\)+25=0 

x\(^2\)+5x-x\(^2\)+25=0

5x+25=0

5x=-25

x=-5

\(x^2-25-7\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+5\right)-7\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)