x(x + 1)(x - 1)(x + 2) = 24

<=> x^4 + 2x^3 - x^2 - 2x = 24

<=> x^4 + 2x^3 - x^2 - 2x - 24 = 0

<=> (x - 2)(x + 3)(x^2 + x + 4) = 0

<=> x - 2 = 0 hoặc x + 3 = 0 hoặc x^2 + x + 4 khác 0

<=> x = 2 hoặc x = -3

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

\(\Leftrightarrow\)\(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\) \(\text{ }\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x=a\), ta có:  \(a\left(a-2\right)=24\)
\(\Leftrightarrow\) \(a^2-2a-24=0\)
\(\Leftrightarrow\) \(\left(a-6\right)\left(a+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}a-6=0\\a+4=0\end{cases}}\) \(\Leftrightarrow\) \(\orbr{\begin{cases}a=6\\a=-4\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x^2+x=6\\x^2+x=-4\end{cases}}\)\(\Leftrightarrow\) \(\orbr{\begin{cases}x^2+x-6=0\\x^2+x+4=0\end{cases}}\)\(\Leftrightarrow\) \(\orbr{\begin{cases}\left(x+3\right)\left(x-2\right)=0\\x^2+x+\frac{1}{4}+\frac{11}{4}=0\end{cases}}\) (1)

Có : \(x^2+x+\frac{1}{4}+\frac{11}{4}=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge0+\frac{11}{4}>0\forall x\) (2)

(1); (2)\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)

Vậy PT có tập nghiệm: S = {-3; 2}

D max thì \(\frac{6}{X^2+2}\)max

mà \(\frac{6}{X^2+2}\) thì X²+2 min   (1)

Ta có X² \(\ge0\)\(\forall X\)

=>X²+2\(\ge2\forall X\)(2)

Từ (1),(2)=> X²+2=2 <=> X =0

Thay X=0 ta có D = 3

Vậy D max =3 <=> X=0

Ta có: x² + 2 \(\ge\)2 \(\forall\) x=> \(\frac{6}{x^2+2}\le\frac{6}{2}=3\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy MaxD = 3 khi x = 0

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

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x};x\ne2;x\ne0\)

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

\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-\left(x-2\right)-2}{x\times\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{x\times\left(x+2\right)-x+2-2}{x\times\left(x-2\right)}=0\)

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

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

\(\Leftrightarrow\frac{x+1}{x-2}=0\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

1) y/(y + 2) - 3/(y - 2) = (y^2 + 8)/(y^2 - 4)

<=> y/(y + 2) - 3/(y - 2) = (y^2 + 8)/((y - 2)(y + 2))

<=> y(y - 2) - 3(y + 2) = y^2 + 8

<=> y^2 - 2y - 3y - 6 = y^2 + 8

<=> y^2 - 5y - 6 = y^2 + 8

<=> -5y - 6 = 8

<=> -5y = 8 + 6

<=> -5y = 14

<=> y = -14/5

2) 7/(2x - 3) + 1/(2x - 2) = 3/(x - 1)

<=> 14(x - 1) + 2x - 3 = 6(2x - 3)

<=> 14x - 14 + 2x - 3 = 12x - 18

<=> 16x - 17 = 12x - 18

<=> 16x - 17 - 12x = -18

<=> 4x - 17 = -18

<=> 4x = -18 + 17

<=> 4x = -1

<=> x = -1/4

a)\(6y\left(y-1\right)=y-1\)

\(6y=\frac{y-1}{y-1}\)

\(6y=1\)

\(y=\frac{1}{6}\)

b)  \(2\left(y+5\right)-y^2-5y=0\)

\(2y+10-y^2-5y=0\)

\(y\left(2-y-5\right)+10=0\)

\(y\left(-3-y\right)=-10\)

\(-3y-2y=-10\)

\(-5y=-10\)

\(y=2\)

c) \(y^3+y=0\)

\(y\left(y^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\y^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y^2=-1\left(vl\right)\end{cases}}}\)

hok tốt!!

Giải pt (1) :(x+3)(2x+1)=0

  =>{x+3=0   /     {2x+1=0

=> {x=-3   /      {x=-1/2 

Để hai pt tương đương thì pt (2) nhận giá trị x=-3 và x=-1/2 .

+)Thay x=-3 vào pt (2) :

     (m-4)(-3)^2 - 2(2m+9)(-3) -4 =0

=> (m-4)9 + 6(2m+9) - 4 = 0

=> 9m - 36+ 12m + 54 - 4= 0

=> 21m + 14 = 0

=> 21m = -14

=> m= -2/3

 Vậy ...

+) Thay x= -1/2 vào pt (2) :

     (m-4)(-1/2)^2 - 2(2m+9)(-1/2) -4 =0

=>1/4(m-4) + 2m +9 - 4 = 0

=>1/4m -1 +2m +9 - 4 =0

=>9/4m +4 =0

=>9/4m = -4 

=>m =-16/9

Vậy ...