1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

Mình thu gọn 2 đa thức trước r mới cộng nhé

\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)

\(P\left(x\right)=\left(3x^2-3x^2\right)+\left(7-4\right)+2x^4-5x+2x^3\)

\(P\left(x\right)=2x^4+2x^3-5x+3\)

\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)

\(Q\left(x\right)=\left(-3x^3+x^3\right)+2x^2+\left(-x^4+5x^4\right)+\left(x+4x\right)-2\)

\(Q\left(x\right)=-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=2x^4+2x^3-5x+3-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=\left(2x^4+4x^4\right)+\left(2x^3-2x^3\right)+\left(-5x+5x\right)+\left(3-2\right)+2x^2\)

\(P\left(x\right)+Q\left(x\right)=6x^4+1+2x^2\)

+) Lỗi nhỏ: Sai ở chỗ: \(\left|x-2+4-3x\right|=\left|-2x-2\right|\)

+) Lỗi lớn: Dấu bằng xảy ra:  \(\hept{\begin{cases}\left(x-2\right)\left(4-3x\right)\ge0\\\left(-2x+2\right)\left(2x-3\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{3}{2}\le x\le1\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le1\)( làm tắt )

Nhưng mà thử vào chọn x= 1=>  A = 3 > 1. Nên bài này sai. 

Làm lại nhé!

A = | x - 2 | + | 2 x - 3  | + | 3  x - 4 |

 = | x - 2 | + | 2 x - 3  | + 3 | x - 4/3 |

= | x -2 | + | x - 4/3 | + | 2x -3 | +2 | x - 4/3 |

= ( | 2 - x | + | x - 4/3 | ) + ( | 3 - 2x  | + | 2x - 8/3 | )

\(\ge\)| 2 -x + x - 4/3 | + | 3 - 2x + 2x -8/3 |

= 2/3 + 1/3 = 1

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(2-x\right)\left(x-\frac{4}{3}\right)\ge0\\\left(3-2x\right)\left(2x-\frac{8}{3}\right)\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{4}{3}\le x\le2\\\frac{4}{3}\le x\le\frac{3}{2}\end{cases}}\Leftrightarrow\frac{4}{3}\le x\le\frac{3}{2}\)

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

`= (2x^4-x^4)+3x^3+(3x^2-2x^2)+(-4x+6x)+2`

`= x^4+3x^3+x^2+2x+2`

`Q(x)=`\(x^4 + 3x^2 + 5x - 1 - x^2 - 3x + 2 + x^3\)

`= x^4+x^3+(3x^2-x^2)+(5x-3x)+(-1+2)`

`= x^4+x^3+2x^2+2x+1`

`P(x)+Q(x)=(x^4+3x^3+x^2+2x+2)+(x^4+x^3+2x^2+2x+1)`

`=x^4+3x^3+x^2+2x+2+x^4+x^3+2x^2+2x+1`

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

`= 2x^4+4x^3+3x^2+4x+3`

`@`\(\text{dn inactive.}\)

P(x)=x^4+3x^3+x^2+2x+2

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

P(x)+Q(x)=2x^4+4x^3+3x^2+4x+3

P(x) = 2x⁴ + 3x³ + 3x² - x⁴ - 4x + 2 - 2x² + 6x

Q(x) = x⁴ + 3x² + 5x - 1 - x² - 3x + 2 + x³

P(x)+Q(x) = 2x⁴ + 3x³ + 3x² - x⁴ - 4x + 2 - 2x² + 6x + x⁴ + 3x² + 5x - 1 - x² - 3x + 2 + x³

P(x)+Q(x) = (2x⁴-x⁴+x⁴) + (3x³+x³) + (3x²-2x²+3x²-x²) - (4x-6x-5x+3x) +(2-1+2)

P(x)+Q(x) = 4x³+3x²-4x+3

Bài 2: 

Ta có: \(P=3x\left(\dfrac{2}{3}x^2-3x^4\right)+9x^2\left(x^3-1\right)+x^2\left(-2x+9\right)-12\)

\(=2x^3-9x^5+9x^5-9x^2-2x^3+9x^2-12\)

=-12

Bài 1: 

a: Ta có: \(x\left(x^2+2\right)+2x\left(1-\dfrac{1}{2}x^2\right)=4\)

\(\Leftrightarrow x^3+2x+2x-x^3=4\)

hay x=1

b: Ta có: \(4x^2\left(x-1\right)+x\left(x^2+4x\right)=40\)

\(\Leftrightarrow4x^3-4x^2+x^3+4x^2=40\)

\(\Leftrightarrow5x^3=40\)

hay x=2

c: Ta có: \(3x\left(x-2\right)-3\left(x^2-3\right)=8\)

\(\Leftrightarrow3x^2-6x-3x^2+9=8\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)

giúp mình với ạ câu nào cũng được

f(x)+g(x)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3-x^4-x^2+x^3-x+5-5x^3+x^2+3x^4

            =(-3x^2-x^2-x^2+x^2)+(x-x)-(1-5)+(x^4+3x^4-x^4+3x^4)-(x^3-2x^3-x^3+5x^3)

            =-4x^2+4+6x^4+3x^3