Phép nhân và phép chia các đa thức

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

\(a,A=x\left(x+5\right)^3:\left(x+5\right)^2+x^2+x\)

\(=x\left(x+5\right)+x^2+x\)

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

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

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

Ta thấy: \(2x\left(x+3\right)⋮x+3\forall x\in Z\)

\(\Rightarrow A⋮x+3\forall x\in Z\)

\(b,B=x^4y^4:y^3x^3+xy+2\)

\(=xy+xy+2\)

\(=2xy+2\)

\(=2\left(xy+1\right)\)

Ta thấy: \(2\left(xy+1\right)⋮xy+1\forall x;y\in Z\)

\(\Rightarrow B⋮xy+1\forall x;y\in Z\)

\(c,C=xy\left(xy+y+1\right)^3:\left(xy+y+1\right)^2+xy\)

\(=xy\left(xy+y+1\right)+xy\)

\(=xy\left(xy+y+1+1\right)\)

\(=xy\left(xy+y+2\right)\)

Ta thấy: \(xy\left(xy+y+2\right)⋮xy+y+2\forall x;y\in Z\)

\(\Rightarrow C⋮xy+y+2\forall x;y\in Z\)

#\(Toru\)

`# \text {DNamNgV}`

`a)`

`A = x(x + 5)^3 \div (x + 5)^2 + x^2 + x`

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

`= x^2 + 5x + x^2 + x`

`= 2x^2 + 6x`

`= 2x(x + 3)`

Vì `2x(x + 3) \vdots (x + 3)`

`=> A=x(x + 5)^3 \div (x + 5)^2 + x^2 + x \vdots x + 3`

`b)`

`B=x^4y^4 \div y^3x^3 + xy + 2`

`= (x^4 \div x^3)(y^4 \div y^3) + xy + 2`

`= xy + xy + 2`

`= 2xy + 2`

`= 2(xy + 1)`

Vì `2(xy + 1) \vdots xy + 1`

`=> B =x^4y^4 \div y^3x^3 + xy + 2 \vdots xy + 1`

`c)`

`C = xy(xy + y + 1)^3 \div (xy + y + 1)^2 + xy`

`= xy(xy + y + 1) + xy`

`= xy(xy + y + 1 + 1)`

`= xy(xy + y + 2)`

Vì `xy(xy + y + 2) \vdots xy + y + 2`

`=> C = xy(xy + y + 1)^3 \div (xy + y + 1)^2 + xy \vdots xy + y + 2.`

4) \(3x^2y^3\cdot A=\dfrac{4}{5}x^4y^5\)

\(\Rightarrow A=\dfrac{4}{5}x^4y^5:3x^2y^3\)

\(\Rightarrow A=\dfrac{4}{15}x^2y^2\)

5) \(-xy^3\cdot A=\dfrac{7}{5}x^2y^6\)

\(\Rightarrow A=\dfrac{7}{5}x^2y^6:\left(-xy^3\right)\)

\(\Rightarrow A=-\dfrac{7}{5}xy^3\)

6) \(\dfrac{3}{4}x^2y^2\cdot A=\dfrac{-5}{6}x^7y^3\)

\(\Rightarrow A=-\dfrac{5}{6}x^7y^3:\dfrac{3}{4}x^2y^2\)

\(\Rightarrow A=-\dfrac{10}{9}x^5y\)

7) \(A\cdot\dfrac{4}{3}x^2y=\dfrac{6}{5}x^3y^5\)

\(\Rightarrow A=\dfrac{6}{5}x^3y^5:\dfrac{4}{3}x^2y\)

\(\Rightarrow A=\dfrac{9}{10}xy^4\)

8) \(-A\cdot\dfrac{1}{2}xy^3=\dfrac{-7}{8}x^3y^6\)

\(\Rightarrow A\cdot\dfrac{-1}{2}xy^3=-\dfrac{7}{8}x^3y^6\)

\(\Rightarrow A=-\dfrac{7}{8}x^3y^6:\left(-\dfrac{1}{2}xy^3\right)\)

\(\Rightarrow A=\dfrac{7}{4}x^2y^3\)

9) \(-A\cdot\left(-4xy\right)^2=\dfrac{6}{7}x^6y^6\)

\(\Rightarrow-A\cdot16x^2y^2=\dfrac{6}{7}x^6y^6\)

\(\Rightarrow-A=\dfrac{6}{7}x^6y^6:16x^2y^2\)

\(\Rightarrow-A=\dfrac{3}{56}x^4y^4\)

\(\Rightarrow A=-\dfrac{3}{56}x^4y^4\)

#Urushi☕

\(=\dfrac{3x^2y^2}{3xy}+\dfrac{6x^2y^3}{3xy}-\dfrac{12xy}{3xy}\)

=xy+2xy^2-4

Lời giải:

$(3x^2y^2+6x^2y^3-12xy):(3xy)$

$=\frac{3x^2y^2}{3xy}+\frac{6x^2y^3}{3xy}-\frac{12xy}{3xy}$

$=xy+2xy^2-4$