a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

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

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

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

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

Bài 1: 

a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)

\(=k^3-64-k^3-128\)

=-192

b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)

\(=8m^3+27n^3-27m^3+8n^3\)

\(=-19m^3+35n^3\)

Bài 4: 

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x=9\)

hay x=1

b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x=7\)

hay \(x=\dfrac{7}{2}\)

$A=x^3-x^2+x+x^2-x^3$

$B=3x^2-6x-x-3x^2$

$C=(x-y)(x^2+xy+y^2)$

$C=x^3-y^3$

$D=3x^3-6x^2-9x-3x^3+2x^2+5x^2-5x$

$=x^2-14x$

a: \(\left(2x^2+5x-2\right)\left(2x^2-4x+3\right)\)

\(=4x^4-8x^3+6x^2+10x^3-20x^2+15x-4x^2+8x-6\)

\(=4x^4+2x^3-18x^2+23x-6\)

b: \(\left(2x-3\right)\left(3x-2\right)-3x\left(2x-5\right)\)

\(=6x^2-4x-9x+6-6x^2+15x\)

=2x+6

c: \(\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

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

\(=x^3-1-x^3-1=-2\)

d: \(\left(x^2+x-1\right)\cdot\left(x^2-x+1\right)=\left(x^2\right)^2-\left(x-1\right)^2\)

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

e: Sửa đề: \(\left(2x+3y\right)^2-\left(2x-3y\right)^2-12xy\)

=(2x+3y+2x-3y)(2x+3y-2x+3y)-12xy

=4x*6y-12xy

=24xy-12xy

=12xy

f: \(\left(x^2-4x\right)\left(5+2x-x^2\right)\)

\(=5x^2+2x^3-x^4-20x-8x^2+4x^3=-x^4+6x^3-3x^2-20x\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cmm

$=[(x+2)-(x+4)][(x+2)+(x+4)]+x^2-3x+1$

$=(-2)(2x+6)+x^2-3x+1$

$=-4x-12+x^2-3x+1$

$A=x^2-7x-11$

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

$=4(x^2+2x+1)-4x^2-8x$

$B=4$

\(A=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\\ A=9x^3-16x-9x^3-72+16x\\ A=-72\)

\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)

\(=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\)

\(=9x^3-16x-9x^3-72+16x=-72\)

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

Ta có:

$(x+4)^2-x^2(x+12)=16$

$x^2+8x+16-x^3-12x^2=16$

$-x^3-11x^2+8x=0$

$-x(x^2+11x-8)=0$

Suy ra:

$x=0$ hoặc $x^2+11x-8=0$

Giải phương trình bậc hai:

$x=\dfrac{-11\pm\sqrt{121+32}}{2}$

$=\dfrac{-11\pm\sqrt{153}}{2}$

$=\dfrac{-11\pm3\sqrt{17}}{2}$

Vậy: $x=0,\quad x=\dfrac{-11+3\sqrt{17}}2,\quad x=\dfrac{-11-3\sqrt{17}}2$