a) \(\left(x+2\right)\left(x^2-24+4\right)\left(x^3+8\right)\)

\(=\left(x+2\right)\left(x^2-20\right)\left(x^3+8\right)\)

\(=\left(x^3-20x+2x^2-40\right)\left(x^3+8\right)\)

\(=x^6+8x^3-20x^4+160x+2x^5+16x^2-40x^3-120\)

\(=x^6+2x^5-20x^4-32x^3+16x^2+160x-120\)

b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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

\(=8x^3-\dfrac{1}{8}\)

c) \(\left(x^2+y\right)\left(x^2-y\right)+y^2+x^4\)

\(=\left(x^2\right)^2-y^2+y^2+x^4\)

\(=x^4-y^2+y^2+x^4\)

\(=2x^4\)

d) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3\)

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

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

\(=27\)

e) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-26x^3\)

\(=\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-26x^3\)

\(=\left(3x\right)^3+y^3-26x^3\)

\(=27x^3+y^3-26x^3\)

\(=x^3+y^3\)

g) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)

\(=\left[x^3+\left(3y\right)^3\right]+\left[\left(3x\right)^3-y^3\right]\)

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3+26y^3\)

Yêu cầu của đề là gì ?

a) Sửa đề:

(x + 2)(x² - 2x + 4)(x³ + 8)

= (x³ + 8)(x³ + 8)

= (x³ + 8)²

b) (2x - 1/2)(4x² + x + 1/4)

= (2x)³ - (1/2)³

= 8x³ - 1/8

c) (x² + y)(x² - y) + y² + x⁴

= (x²)² - y² + y² + x⁴

= 2x⁴

d) (x + 3)(x² - 3x + 9) - x³

= x³ + 3³ - x³

= 27

e) (3x + y)(9x² - 3xy + y²) - 26x³

= (3x)³ + y³ - 26x³

= 27x³ + y³ - 26x³

= x³ + y³

g) (x + 3y)(x² - 3xy + 9y²) + (3x - y)(9x² + 3xy + y²)

= x³ + (3y)³ + (3x)³ - y³

= x³ + 27y³ + 27x³ - y³

= 28x³ + 26y³

giúp mình vớiiii

c)  \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

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

d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)

e)  \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)

Đường Quỳnh Giang ơi

Giari giúp mình bài này

1.Tìm a,b để x^10 + ax^3 + b chia hết x^2-1 dư 2x+1

2.Tìm a,b để đa thức x^3 +ax +b chia hết x^2-x-2

Mong các bạn làm giúp mình, mình rất gấp

Cảm ơn nha.

\(\sqrt{4-x^2}=\sqrt{x+2}\) (ĐK: \(-2\le x\le2\))

\(\Leftrightarrow4-x^2=x+2\)

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

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

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

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

_______

\(\sqrt{9x^2-4}=2\sqrt{3x-2}\) (ĐK: \(x\ge\dfrac{2}{3}\)) 

\(\Leftrightarrow9x^2-4=4\left(3x-2\right)\)

\(\Leftrightarrow9x^2-4=12x-8\)

\(\Leftrightarrow9x^2-12x+4=0\)

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

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)

b) \(x^3-6x^2+9x=0\)

\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)

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

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

Vậy \(x=0\)hoặc \(x=3\)

a. ( x - 1 )³ + 1 + 3x ( x - 4 ) = 0

<=> x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

<=> x³ - 9x = 0

<=> x ( x² - 9 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b. x³ - 6x² + 9x = 0

<=> x ( x² - 6x + 9 ) = 0

<=> x ( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a) ( x - 1 )³ + 1 + 3x( x - 4 ) = 0

⇔ x³ - 3x² + 3x - 1 - 1 + 3x² - 12x = 0

⇔ x³ - 9x = 0

⇔ x( x² - 9 ) = 0

⇔ \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

b) x³ - 6x² + 9x = 0

⇔ x( x² - 6x + 9 ) = 0

⇔ x( x - 3 )² = 0

⇔ \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

\(9x^2-1+\left(3x-1\right).\left(x+2\right)=0\)

\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)

\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)

\(\Leftrightarrow12x^2+5x=3\)

\(\Leftrightarrow12x^2+5x-3=0\)

\(\Leftrightarrow12x^2-4x+9x-3=0\)

\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)

\(\Leftrightarrow\left(4x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)

( 3x - 1 )( x + 3 ) + 9x² - 1 = 0

<=> 3x² + 9x - x - 3 + 9x² - 1 = 0

<=> 12x² + 8x - 4 = 0

<=> 4( 3x² + 2x - 1 ) = 0

<=> 3x² + 2x - 1 = 0 

<=> 3x² + 3x - x - 1 = 0

<=> ( 3x² + 3x ) - ( x + 1 ) = 0

<=> 3x( x + 1 ) - 1( x + 1 ) = 0

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

<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Vậy S = { 1/3 ; -1 }

\(\frac{x+1}{3}>\frac{3x-2}{5}\)

\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Leftrightarrow5x+5>9x-6\)

\(\Leftrightarrow5x-9x>-6-5\)

\(\Leftrightarrow-4x>-11\)

\(\Leftrightarrow x< \frac{11}{4}\)

Bài làm:

a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)

\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)

\(\Leftrightarrow12x^2+8x-4=0\)

\(\Leftrightarrow3x^2+2x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)

b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Rightarrow5x+5>9x-6\)

\(\Leftrightarrow4x< 11\)

\(\Rightarrow x< \frac{11}{4}\)

(3x-1)(x+3)+9x²-1=0

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\-2x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

b) \(x+\frac{1}{3}>3x-\frac{2}{5}\)

\(\Leftrightarrow x+\frac{1}{3}-3x+\frac{2}{5}>0\)

\(\Leftrightarrow-2x+\frac{11}{15}>0\)

\(\Leftrightarrow2x< \frac{11}{15}\Leftrightarrow x< \frac{22}{15}\)

Có : 

b) (x - 8)(x + 8) = (x - 4)(x² + 4x + 16)

  x² - 8² = x³ - 4³

x² - 2^6 - x³ + 2⁶  = 0

x² . ( x - 1 ) = 0

x²  = 0 hoặc x-1 = 0

x= 0 hoặc x = 1

 Vâỵ....

b)=x^3+x^2+4x^2+4x+4x+4=x(x+1)+4x(x+1)+4(x+1)=(x+1)(x+4x+4)

Thiên Ân ơi, bạn giải giúp mình câu c được kh

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

\(\Rightarrow x^3-3x^2-3x-1\) chia hết \(x^2+x+1\) khi \(3⋮x^2+x+1\)

\(\Rightarrow x^2+x+1=Ư\left(3\right)\) (1)

Mà x nguyên dương \(\Rightarrow x^2+x+1\ge1^2+1+1=3\) (2)

(1);(2) \(\Rightarrow x^2+x+1=3\)

\(\Rightarrow x=1\)