\(a.\)

\(\left(x-8\right)\left(x^3+8\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(S=\left\{8,-2\right\}\)

\(b.\)

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

\(\Leftrightarrow4x-3-x-5-30+3x=0\)

\(\Leftrightarrow6x-38=0\)

\(\Leftrightarrow x=\dfrac{38}{6}\)

\(S=\left\{\dfrac{38}{6}\right\}\)

a) \(\left(x-8\right)\left(x^3+8\right)=0\)

=>\(x-8=0 => x=8\)

hoặc \(x^3+8=0\)=>\(x=-2\)

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

\(< =>3x-8=3\left(10-x\right)\)

\(< =>3x-8-30+3x=0\)

\(< =>6x=38=>x=\dfrac{38}{6}=\dfrac{19}{3}\)

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` 

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

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

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

`=> 3x-10=30-3x`

`=> 6x=40`

`=> x=20/3`

Vậy `x=20/3` 

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: \(x\left(x-3\right)+2x-6=0\)

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

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

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

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

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

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}`.

\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)

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

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

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`

\(a.\left(x-8\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\\left(x+2\right)\left(x^2-2x+4\right)=0\end{matrix}\right.\)

Ta có: \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3\ge3>0\)
\(\Rightarrow x=-2\)

Vậy \(S=\left\{-2;8\right\}\)

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

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

\(\Leftrightarrow6x=38\)

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

Vậy \(S=\left\{\dfrac{19}{3}\right\}\)

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

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

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

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

<=> (x+3)(x+2)=0

TH1 x+3=0 <=> x=-3

TH2 x+2=0 <=> x=-2

Vậy.... 

a,(2x-5^2)-4x(x-3)=0

=> 2x-25-4x²+12x=0

=>-4x²+14x-25=0

đề bài ý a sai nha

b, 6x²-7x=0

=>x(6x-7)=0

=>x=0 và 6x-7=0

=>x=0 và x=7/6

vậy x=0 và x=7/6

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

a)5x.(x+1)-5.(x+1).(x-2)=0

⇒5x(x+1)-(5x-10)(x+1)=0

⇒(x+1)(5x-5x+10)=0

⇒10(x+1)=0

⇒x+1=0⇒x=-1

a) \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow5\left(x+1\right)\left(x-x+2\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

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

b) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)^2=4\)

\(\Leftrightarrow4x^2-7x-2-4x^2+12x-9=4\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)