Phép nhân và phép chia các đa thức

Bài 3:

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

\(=3x^3+9x^2y-4x-12y\)

2) \(\left(x+3\right)\left(x^2+3x\right)\)

\(=x^3+3x^2+3x^2+9x\)

\(=x^3+6x^2+9x\)

3) \(\left(xy-1\right)\left(xy+5\right)\)

\(=x^2y^2+5xy-xy-5\)

\(=x^2y^2+4xy-5\) 

4) \(\left(3x+5y\right)\left(2x-7y\right)\)

\(=6x^2-21xy+10xy-35y^2\)

\(=6x^2-11xy-35y^2\)

5) \(-\left(x-1\right)\left(-x^2+2y\right)\)

\(=\left(x-1\right)\left(x^2-2y\right)\)

\(=x^3-2xy-x^2+2y\)

6) \(\left(-x^2+2y\right)\left(x^2+2y\right)\)

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

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

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

7) \(\left(x+3y\right)\left(x-3y+2\right)\)

\(=x^2-3xy+2x+3xy-9y^2+6y\)

\(=x^2+2x-9y^2+6y\)

8) \(\left(x+2y\right)\left(x-2y+3\right)\)

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

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

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

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

10) \(\left(x^2-xy+y^2\right)\left(x+y\right)\)

\(=x^3+y^3\) 

Bài 16:

a) \(A-B+C\)

\(=\left(5xy^2-4x^2y-6x^2\right)-\left(8x^2y-4xy^2+3y^2\right)+\left(-2xy^2+3x^2y+5x^2\right)\)

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

\(=\left(5xy^2+4xy^2-2xy^2\right)-\left(4x^2y+8x^2y-3x^2y\right)-\left(6x^2-5x^2\right)-3y^2\)

\(=7xy^2-9x^2y-x^2-3y^2\)

b) \(2\left(A+B\right)+C\)

\(=2\cdot\left(5xy^2-4x^2y-6x^2+8x^2y-4xy^2+3y^2\right)+\left(-2xy^2+3x^2y+5x^2\right)\)

\(=2\cdot\left(xy^2+4x^2y-6x^2+3y^2\right)-2xy^2+3x^2y+5x^2\)

\(=2xy^2+8x^2y-12x^2+6y^2-2xy^2+3x^2y+5x^2\)

\(=11x^2y-7x^2+6y^2\)

`#3107.101107`

`K = -x^2 + 10x - 30`

`= -(x^2 - 10x + 30)`

`= -[ (x^2 - 2*x*5 + 5^2) + 5]`

`= -[ (x - 5)^2 + 5]`

`= -(x - 5)^2 - 5`

Vì `-(x - 5)^2 \le 0` `AA` `x`

`=> -(x - 5)^2 - 5 \ge -5` `AA` `x`

Vậy, `K_(\text {max})` là `-5` khi `(x - 5)^2 = 0`

`<=> x - 5 = 0`

`<=> x = 5`

_____

`Q = -25y^2 - 20y - 2`

`= -(25y^2 + 20y + 2)`

`= - [ (5y)^2 + 2*5y*2 + 2^2 - 2]`

`= - [(5y + 2)^2 - 2]`

`= - (5y + 2)^2 + 2`

Vì `-(5y + 2)^2 \le 0` `AA` `y`

`=> -(5y + 2)^2 + 2 \ge 2` `AA` `y`

Vậy, `Q_(\text {max})` là `2` khi `(5y + 2)^2 = 0`

`<=> 5y + 2 = 0`

`<=> 5y = -2`

`<=> y = -2/5.`

`#3107.101107`

a)

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

`<=> 9 + 6x + x^2 - (x^2 - 1) = 1`

`<=> 9 + 6x + x^2 - x^2 + 1 = 1`

`<=> 6x + 10 = 1`

`<=> 6x = 1 - 10`

`<=> 6x = -9`

`<=> x = -3/2`

Vậy, `x = -3/2`

b)

`(9 + x)(9 - x) + x(x - 7) = 11`

`<=> 81 - x^2 + x^2 - 7x = 11`

`<=> 81 - 7x = 11`

`<=> 7x = 70`

`<=> x = 10`

Vậy, `x = 10`

c)

`(2x - 3)(2x + 3) - 4(x - 1)^2 = 19`

`<=> 4x^2 - 9 - 4(x^2 - 2x + 1) = 19`

`<=> 4x^2 - 9 - 4x^2 + 8x - 4 = 19`

`<=> 8x - 13 = 19`

`<=> 8x = 32`

`<=> x = 4`

Vậy, `x = 4.`

______

Sử dụng các hđt:

1. `A^2 - B^2 = (A - B)(A + B)`

2. `(A - B)^2 = A^2 - 2AB + B^2`

3. `(A + B)^2 = A^2 + 2AB + B^2`

d)

Đề đâu ah .

Chắc l: `(n+2)*(n^2+3n-1)-n^3+2⋮5` nhỉ?

Ta có : `(n+2)*(n^2+3n-1)-n^3+2`

`=n*n^2+n*3n-n*1+2*n^2+2*3n-2*1-n^3+2`

`= n^3 +3n^2 -n +2n^2+6n-2-n^3+2`

`= 5n^2 +5n`

`=5n(n+1)`

Vì `5n` luôn chia hết cho `5`

`=> 5n(n+1)⋮5`

\(E=6a+3-9a^2\)

\(=-\left(9a^2-6a-3\right)\)

\(=-\left(\left(3a\right)^2-2.3a+1-4\right)\)

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

Vì \(-\left(3a-1\right)^2\le0\forall a\)

\(\Rightarrow E\le4\forall x\)

\(MaxE=4\Leftrightarrow x=\dfrac{1}{3}\)

\(e=6a+3-9a^2\)

\(=-9a^2+6a-1+4\)

\(=-\left(9a^2-6a+1\right)+4\)

\(=-\left[\left(3a\right)^2-2\cdot3a\cdot1+1^2\right]+4\)

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

Ta thấy: \(\left(3a-1\right)^2\ge0\forall a\)

\(\Rightarrow-\left(3a-1\right)^2\le0\forall a\)

\(\Rightarrow-\left(3a-1\right)^2+4\le4\forall a\)

Dấu \("="\) xảy ra \(\Leftrightarrow3a-1=0\Leftrightarrow a=\dfrac{1}{3}\)

Vậy \(e_{max}=4\) khi \(a=\dfrac{1}{3}.\)

#\(Toru\)

\(P\times Q\)

\(=\left(2x+3y\right)\times\left(2x-3y\right)\)

\(=\left(2x\right)^2-\left(3y\right)^2\)

\(=4x^2-9y^2\)

viết rõ đề ra nha bạn để mọi người hỗ trợ 

\(P.Q=\left(2x-3y\right)\left(2x+3y\right)\)

         \(=4x^2-9y^2\)

Ta có: \(A=x^2y^4+2x^3y^3\) 

Để A chia hết cho \(B=x^ny^3\) thì:

\(\left\{{}\begin{matrix}2x^3y^3⋮x^ny^3\\x^2y^4⋮x^ny^3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^3⋮x^n\\x^2⋮x^n\end{matrix}\right.\)

\(\Rightarrow x^0\le x^n\le x^2\)

\(\Rightarrow0\le n\le2\) 

Bài 23:

\(P=x^5y^2-3x^3y+5-x^3y+ax^5y^2=\left(1+a\right)x^5y^2-4x^3y+5\\ Ta.có:x^5y^2\Rightarrow bậc.7\\ -4x^3y\Rightarrow Bậc.4\\ 5:Bậc.0\\ Vậy:P.đa.thức.bậc.4\Leftrightarrow1+a=0\Leftrightarrow a=-1\\ Vậy:a=-1\)

\(Bài.22:A=x^2-3xy-y^2+1;B=-2x^2+y^2-7xy-5;C=-4+x^2+10xy\\ a,A+B+C=\left(x^2-3xy-y^2+1\right)+\left(-2x^2+y^2-7xy-5\right)+\left(-4+x^2+10xy\right)\\ =\left(x^2-2x^2+x^2\right)+\left(y^2-y^2\right)-\left(7xy+3xy-10xy\right)+1-5-4=-8\)

Vậy: A+B+C không phụ thuộc gì vào biến

\(b,C+A-B=0\\ \Leftrightarrow C=B-A=\left(-2x^2+y^2-7xy-5\right)-\left(x^2-3xy-y^2+1\right)\\ =\left(-2x^2-x^2\right)+\left(y^2+y^2\right)-\left(7xy-3xy\right)-\left(5+1\right)\\ =-3x^2+2y^2-4xy-6\)

Bài 20:

Bậc của đơn thức là 25 

<=> (n+2)+1+3+2 + (n-1)+4=25

<=> 2n+11=25

<=>2n=25-11

<=>2n=14

<=>n=14:2=7

Vậy n=7