x = 0 nha bạn

\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

đặt \(x^2+5x+5=t\)

\(\Leftrightarrow t^2-25=0\Rightarrow\left\{{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

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

a)   ta có :(x-1)(x-2)(x+3)(x+4)=24

           <=>[(x-1)(x+3)].[(x-2)(x+4)] =24

          <=>(x^2 +2x -3)(x^2+2x -8)=24

         đặt x^2  +2x -3  =a =>  (x^2 +2x -3)(x^2 +2x-8)=a(a-5) =24

                                                                                 <=>a^2 -5a-24=0

                                                                                <=>(a-8)(a+3)=0  <=> a-8=0 hoặc a+3=0 <=>a=8 hoặc a=-3

+) với a=8 => x^2 +2x-3=8 <=>x^2 +2x-11=0<=>(x+1)^2 -10=0   (vô nghiệm)  vì (x+1)^2  >=0

+) với a=-3=>x^2 +2x-3=-3<=>x^2 +2x=0<=>x.(x+2)=0  <=> x=0 hoặc x+2=0 <=>x=0 hoặc x=-2

Vậy tập nghiệm của pt là S={0;-2}

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt   \(x^2+5x+4=a\)ta có:

                  \(a\left(a+2\right)-24\)

              \(=a^2+2a+1-25\)

              \(=\left(a+1\right)^2-25\)

              \(=\left(a-4\right)\left(a+6\right)\)

Thay trở lại ta được:

                  \(\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(a.x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\) ( ĐKXĐ : \(x\ne0\) )

\(\Leftrightarrow x^2+\dfrac{1}{x^2}-x-\dfrac{1}{x}=0\Leftrightarrow\left(x^2-\dfrac{1}{x}\right)+\left(\dfrac{1}{x^2}-x\right)=0\)

\(\Leftrightarrow-x\left(\dfrac{1}{x^2}-x\right)+\left(\dfrac{1}{x^2}-x\right)=0\Leftrightarrow\left(\dfrac{1}{x^2}-x\right)\left(1-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\\dfrac{1}{x^2}-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\1-x^3=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\Leftrightarrow x=1\) ( x² + x + 1 loại nhé nếu phân tích ra thì ta được \(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\) )

Vậy \(S=\left\{1\right\}\)

b, \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow x\left(x+3\right).\left(x+1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)-24=0\)

\(\Leftrightarrow\left(x^2+3x+1\right)-1-24=0\Leftrightarrow\left(x^2+3x+1\right)-25=0\)

\(\Leftrightarrow\left(x^2+3x+1-5\right)\left(x^2+3x+1+5\right)=0\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\end{matrix}\right.\)

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

Vậy \(S=\left\{-4;1\right\}\)

e, \(\left(x^2+x+1\right)-2x^2-2x=5\Leftrightarrow\left(x^2+x+1\right)-2x^2-2x-2-3=0\)

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

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x-1\right)-3=0< =>\left(x^2+x\right)^2-4=0\) 

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

\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\) ( x^2 + x + 2 loại nhé y như mấy câu trên luôn khác 0 ! )

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

Vậy \(S=\left\{-2;1\right\}\)

\(\left(x+1\right).\left(x+1\right)\left(x+2\right)=24\)

\(\left(x^2+2x+1\right).\left(x+2\right)=24\)

\(x^3+2x^2+2x^2+4x+x+2=24\)

\(x^3+4x^2+5x+2=24\)

rồi đặt nhân tử chung

bạn tự giải nhé

=>-(x+3)^2*(x-4)(x+12)=x^2-48x+576

=>-(x^2+6x+9)(x^2+8x-48)=x^2-48x+576

=>-x^4-14x^3-9x^2+216x+432=x^2-48x+576

=>x^4+14x^3+10x^2-264x+144=0

=>(x^2+4x-24)(x^2+10x-6)=0

=>\(x\in\left\{-5+\sqrt{31};-5-\sqrt{31};-2+2\sqrt{7};-2-2\sqrt{7}\right\}\)

bài nào ạ

À vậy  để c đăng lại nha.

ĐK:...

\(\Leftrightarrow\dfrac{2x+3}{x-3}-\dfrac{4}{x+3}-\dfrac{24}{\left(x-3\right)\left(x+3\right)}-2=0\)

\(\Leftrightarrow\dfrac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)-24-2\left(x^2-9\right)}{x^2-9}=0\)

\(\Leftrightarrow2x^2+6x+3x+9-4x+12-24-2x^2+18=0\)

\(\Leftrightarrow5x+15=0\)

<=> x = -3 (ko t/m đk)

=> Pt vô nghiệm