a) (x - 8 )( x³ + 8) = 0

\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b)(4x - 3) – ( x + 5) = 3(10 - x)

\(\Leftrightarrow4x-3-x-5=30-3x\)

\(\Leftrightarrow3x-8=30-3x\)

\(\Leftrightarrow3x-8-30+3x=0\)

\(\Leftrightarrow6x-38=0\)

\(\Leftrightarrow x=\dfrac{19}{3}\)

Sửa lại câu `b) :` 

`a)`

`( x-8 )( x^3 + 8 )`

`=> x-8=0` hoặc `x^3+8=0`

`=> x=8` hoặc `x^3 = -8=(-2)^3`

`=> x=8` hoặc `x=-2`

Vậy `x in { -2;8}`

`b)`

`( 4x-3 ) - ( x+5) = 3( 10-x)`

`=> 4x-3-x-5=30-3x`

`=> ( 4x-x)+(-3-5)=30-3x`

`=> 3x-8=30-3x`

`=> 6x=38`

`=> x=19/3`

Vậy `x=19/3` 

Bài 2 

P(x) + Q(x) =  x³ – 6x + 2 + 2x² - 4x³ + x - 5 =  - 3x³ + 2x² – 5x - 3 

P(x) - Q(x) = x³ – 6x + 2 - 2x² + 4x³ - x + 5 = 5x³ − 2x² − 7x+7

Bai 3

a)(x-8)(x³+8)=0

=>x-8=0 hoac x³+8=0

=>x   =8 hoac x³    =-8

=>x   =8 hoac x     =-2

Vậy x=8 hoặc x=-2

b)(4x-3)-(x+5)=3(10-x)

=>4x-3-x-5=30-3x

=>4x-x+3x=30+3+5

=>x(4-1+3)=38

=>6x         =38

=>x           =\(\dfrac{38}{6}\)

=>x           =\(\dfrac{19}{3}\)

Vậy x=\(\dfrac{19}{3}\)

a) `(x-8)(x^3+8)=0`

`<=>(x-8)(x+2)(x^2-2x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\) (Vì `x^2-2x+4 \ne 0 forall x)`

Vậy `A={8;-2}`.

b) `(4x-3)-(x+5)=3(10-x)`

`,=>4x-3-x-5=30-3x`

`<=>3x-8=30-3x`

`<=>6x=38`

`<=>x=19/3`

Vậy `S={19/3}`.

\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

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

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

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)

\(a,\left(x-8\right)\left(x^3+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)

TK

`a.(x-8)(x+8)=0`

`⇔³{x−8=0x³+8=2 `

`⇔³³{x=8x³=−2³ `

`⇔{x=8x=−2`

Vậy ` x = 8;-2`

`b. ( 4 x − 3 ) − ( x + 5 ) = 3 . ( 10 − x )`

`⇔ 4 x − 3 − x − 5 = 30 − 3 x`

`⇔ 3 x − 8 = 30 − 3 x`

`⇔ 3 x + 3 x = 30 + 8`

`⇔ 6 x = 38`

`⇔ x = 19/ 3`

Vậy ` x = 19/ 3`

Bài 3:

\(\left(x-8\right)\left(x3+8\right)=0\) 

\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x3+8=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{-8}{3}\end{matrix}\right.\) 

\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\) 

      \(4x-3-x-5=30-3x\)  

          \(4x-x+3x=30+3+5\) 

                         \(6x=38\) 

                           \(x=38:6\) 

                           \(x=\dfrac{19}{3}\)

ảnh lỗi r ạ

Lỗi rùi

a: \(x\left(x-3\right)+2x-6=0\)

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

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

b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)

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

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

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