`a, 504 : (16-3x)=72`

`<=> 16-3x=7`

`<=>3x=9`

`<=>x=3`

`b,272-(4x+15)=45`

`<=>4x+15=227`

`<=>4x=212`

`<=>x=53`

a: =>16-3x=504:72=7

=>3x=9

=>x=3

b: =>4x+15=272-45=227

=>4x=212

=>x=53

a) 504 : (16 - 3\(\times\)x) = 72

=> 16 - 3\(\times\)x = 7

=> 3\(\times\)x = 9

=> x = 3

d) 272 - (4\(\times\)x + 15) = 45

=> 4\(\times\)x + 15 = 227

=> 4\(\times\)x = 212

=> x = 53

b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)

\(\Rightarrow x^3-1-x\left(x^2-9\right)=8\)

\(\Rightarrow x^3-1-x^3+9x=8\)

\(\Rightarrow9x=9\Rightarrow x=1\)

c) \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)\left(x^2+4x+4\right)=-16\)

\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)\left(x+2\right)^2=-16\)

\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)^3=-16\)

\(\Rightarrow x^3-4x^2+2x-8-\left(x^3+6x^2+12x+8\right)=-16\)

\(\Rightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Rightarrow-10x^2-10x-16=-16\)

\(\Rightarrow10x^2+10x=0\)

\(\Rightarrow10x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Sửa đề : 

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)

\(\Leftrightarrow x=2018-1\)

\(\Leftrightarrow x=2017\)

Vậy ...

Sửa đề  \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2018}\)

\(\Leftrightarrow x=2017\)

x = 2017 nha bạn 

b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)

Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)

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

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)

b) 3x(x\(^3\) +12x-2x\(^2\)-8=0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
 

a) x\(^2\)+5x=0

x(x+5)=0

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

c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36

x\(^3\)-27+x(x+2)(2-x)=36

4x-27=36

4x=36+27

4x=63

x=\(\dfrac{63}{4}\)

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

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

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 \(\in\left\{3\right\}\)

bbbbbbbbbbbbffv

Đề là $x(x+3)^3$ hay $x(x+3)^2$ hả bạn?

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x+7=16\)

\(\Leftrightarrow9x=9\)

hay x=1