\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\)

* x² - 2x - 3 = x²- 3x + x - 3 = x(x-3 ) + ( x - 3) = ( x - 3 ) (  x + 1 )

\(\Leftrightarrow\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm3;x\ne-1\right)\)

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

\(\Leftrightarrow x^2-2x+1+8x+24=2x^2+6x\)

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

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

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

Vậy \(S=\left\{-5;5\right\}\)

=>x^3+6x^2+12x+8+1/3(8x^3-24x^2+24x-8)=1/5x+2/5+8

=>x^3+6x^2+12x+8+8/3x^3-8x^2+8x-8/3=1/5x+42/5

=>11/3x^3-2x^2+20x+16/3-1/5x-42/5=0

=>11/3x^3-11/5x^2+20x-46/15=0

=>\(x\simeq0,16\)

giup minh với gần thi rùi

\(pt \Leftrightarrow x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8\)

\(\Leftrightarrow 18x^3-33x^2-57x-18=0\)

\(\Leftrightarrow (3x+2)(6x^2-15x-9)=0\)

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

\(\Leftrightarrow x\in\{\dfrac{-1}{2},\dfrac{-2}{3},3\}\)

a: \(=\dfrac{35x^3-14x^2+55x^2-22x+35x-14+9}{5x-2}\)

\(=7x^2-11x+7+\dfrac{9}{5x-2}\)

b: \(=\dfrac{\left(2x-3\right)\left(4x^2+6x+9\right)}{2x-3}=4x^2+6x+9\)

a: Đặt x-3=a; x+1=b

Theo đề, ta có: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>(x-3)(x+1)(2x-2)=0

hay \(x\in\left\{3;-1;1\right\}\)

b: \(\Leftrightarrow\left(2x^2+1\right)^2+2x\left(2x^2+1\right)-15x^2-9x^2=0\)

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

\(\Leftrightarrow\left(2x^2+1\right)^2+6x\left(2x^2+1\right)-4x\left(2x^2+1\right)-24x^2=0\)

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

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

\(\Leftrightarrow x^2+3x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x^2+3x+\dfrac{9}{4}=\dfrac{7}{4}\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{4}\)

hay \(x\in\left\{\dfrac{\sqrt{7}-3}{2};\dfrac{-\sqrt{7}-3}{2}\right\}\)

 x^3 + (x - 2)(2x + 1) = 8

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

<=> x^3 + 2x^2 - 3x - 10 = 0

<=> (x - 2)(x^2 + 4x + 5) = 0

vì x^2 + 4x + 5 > 0 nên:

<=> x - 2 = 0

<=> x = 2

       x^3+(x-2)(2x+1)=8

<=>x^3+2x^2-3x-10=0

<=>x^3-2x^2+4x^2-8x+5x-10=0

<=>x^2(x-2)+4x(x-2)+5(x-2)=0

<=>(x-2)(x^2+4x+5)=0

Mà x^2+4x+5>0

=>x-2=0<=>x=2

   Hok tốt !

(x-1)\(^3\)+ (2x+3)\(^3\)= (3x+2)\(^3\)

Đặt x-1 = a (a thuộc N*) (1)

2x + 3 =b ( b thuộc N*) (2)

=> (x-1) + (2x+3) = 3x+2

Ta có a\(^3\)+ b\(^3\)=( a+b)\(^3\)

=> a\(^3\) + b\(^3\)= a\(^3\)+ 3a\(^2\)b + 3ab\(^2\)+ b\(^3\)

=> 3a\(^2\)b + 3ab\(^2\)=0

=> 3ab(a+b) = 0

=> a=0 hoặc b = 0

+) Thay a=0 vào (1), ta có: x-1=0 <=> x=1

+) Thay b=0 vào (2) ta có 2x+3 =0 <=> x=\(\dfrac{-3}{2}\)

Vậy nghiệm của pt là 1; \(\dfrac{-3}{2}\)

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:

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