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

Thank bạn <3

ta có:\(x^3+x^2+2x^2+2x+2x+2=0\)0

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

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

Do \(x^2+2x+2\ne0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

vậy phương trình trên có tập nghiệm là :S=(-1) 

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tớ chịu.

hi hi.

bạn viết rõ đề ra nhé

b, \(\left|4x-8\right|=1-x\)ĐK : \(x\le1\)

TH1 : \(4x-8=1-x\Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\)( ktm )

TH2 : \(4x-8=x-1\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\)( ktm )

b) Ta có: \(\left|4x-8\right|=1-x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-8=1-x\left(x\ge2\right)\\4x-8=x-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+x=1+8\\4x-x=-1+8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=9\\3x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(loại\right)\\x=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)

(3x-4-x-1)(3x-4+x+1)=0
(2x-5)(4x-3)=0
2x-5 = 0 hoặc 4x-3=0
2x=5      hoặc 4x=3
x=5/2     hoặc   x=3/4

(3x - 4 - x - 1)(3x - 4 + x + 1) = 0

(2x - 5)(4x - 3) = 0

2x - 5 = 0           hoặc               4x - 3 = 0

x = 5/2               hoặc               x = 3/4

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(8-5x)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\) 

Vậy `S={-3,8/5}`

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(-5x+8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)

Vậy `S={-3;8/5}`.

= x^2(X-1) - 4(x^2-2x+1)

=x^2(x-1)-4(x-1)^2

=(x-1)(x^2-4x+4)

=(x-1)(x-2)^2

a)    x³-x²-21x+45=0

<=> x³+5x²-6x²-30x+9x+45=0

<=> (x+5)(x²-6x+9)=0

<=> (x+5)(x²-3x-3x+9)=0

<=> (x+5)(x-3)²=0

 Vậy S={-5;3}

b)    X³+3X²+4X+2=0

<=>  X³+X²+2X²+2X+2X+2=0

<=> (X+1)(X²+2X+2)=0

VÌ  X²+2X+2 >=0

NÊN S={-1}

C)    X⁴+7X-8=0

<=> X⁴-X³+X³-X²+X²-X+8X-8=0

<=> (X-1)(X³+X²+X+8)=0

VÌ X³+X²+X+8>=0

NÊN S={1}

D)     6X⁴-X³-7X²+X+1=0

<=>  6X⁴-6X³+5X³-5X²-2X²+2X-X+1=0

<=>  (X-1)(6X³+5X²-2X-1)=0

<=> (X-1)(6X³-3X²+8X²-4X+2X-1)=0

<=> (X-1)(2X-1)(3X^2_4X+1)=0

<=>  (X-1)(2X-1)(3X²-3x-x+1)=0

<=> (X-1)²(2X-1)(3x-1)=0

vậy S={1/3;1/2;1}