Ta có (2x³ - x² + 4x + a) : (x + 2) = 2x² + 3x - 2 dư a + 4

=> (2x³ - x² + 4x + a) \(⋮\)(x + 2) <=> a + 4 = 0 => a = -4

Vậy a = -4 thì  (2x³ - x² + 4x + a) \(⋮\)(x + 2)

Để 2 đa thức chia hết cho nhau 

\(\Leftrightarrow a-28=0\Leftrightarrow a=28\)

Vậy 

 (6x^3 - 7x^2 - x - 2): (2x + 1) = 3x^2 - 5x - 2.

dn8qJmB.png (1366×768) vào thống kê của mình xem tính chia nhó !!!

Vậy \(\left(6x^3-7x^2-x-2\right):\left(2x+1\right)=3x^2-5x+2\)( dư -4 )

Hoc tot 

a) 10x + 15y = 5(2x + 3y)

b) x² - 2xy - 4 + y²

= (x² - 2xy + y²)  - 4

= (x - y)² - 2²

= (x - y + 2)(x - y - 2)

c) x(x + y) - 3x - 3y

= x(x + y) -3(x + y)

= (x - 3)(x + y)

a, \(10x+15y=5\left(2x+3y\right)\)

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

c, \(x\left(x+y\right)-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\)

Ta có (a + b + c)² = a² + b² + c²

=> a² + b² + c² + 2ab + 2bc + 2ca = a² + b² + c²

=> 2(ab + bc + ca) = 0

=> ab + bc + ca = 0

=> \(\frac{abc}{c}+\frac{abc}{a}+\frac{abc}{b}=0\)

=> \(abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\left(\text{Vì }abc\ne0\right)\)

=> \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{a^2b}+\frac{3}{ab^2}=-\frac{1}{c^3}\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)(Vì 1/a + 1/b = -1/c)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)(đpcm)

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

a, Thay x = -3 ta được : 

\(=\left(-3+1\right)^{-3}-2\left[\left(-3\right)^2-3-2\right]+2\)

\(=-\frac{1}{8}-8+2=-\frac{1}{8}-\frac{64}{8}+\frac{16}{8}=\frac{-49}{8}\)

b, Ta có : \(m=0\)hay \(\left(x+1\right)^x-2\left(x^2+x-2\right)+2=0\)

... =))?