Phép nhân và phép chia các đa thức

4:

a: \(=5x^2y^2\cdot\dfrac{7}{10}x^4y\cdot40x^2z^3\)

\(=\left(5\cdot\dfrac{7}{10}\cdot40\right)\left(x^2y^2\cdot x^4y\cdot x^2z^3\right)\)

\(=140x^8y^3z^3\)

Bậc là 8+3+3=14

b: \(=-\dfrac{1}{2}\cdot ab\cdot\dfrac{-4}{3}\cdot a^2bc\cdot5c^2b^3\)

\(=\left(\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot5\right)\cdot ab\cdot a^2b\cdot c\cdot c^2b^3\)

\(=\dfrac{10}{3}a^3b^5c^3\)

Bậc là 11

c: \(=-1.2\cdot ab\cdot100a^4b^2\cdot c^3\cdot\left(-1.5\right)\cdot a^2c\)

\(=1.8\cdot100\cdot ab\cdot a^4b^2c^3\cdot a^2c\)

\(=180a^7b^3c^4\)

Bậc là 14

a) \(2x^2+A=4,5x^2\)

\(\Leftrightarrow A=4,5x^2-2x^2\)

\(\Leftrightarrow A=2,5x^2\)

b) \(-2,5m^2n+B=5m^2n\)

\(\Leftrightarrow B=5m^2n+2,5m^2n\)

\(\Leftrightarrow B=7,5m^2n\)

c) \(C-3,2ab^3=-11ab^3\)

\(\Leftrightarrow C=-11ab^3+3,2ab^3\)

\(\Leftrightarrow C=-7,8ab^3\)

d) \(-xyz^2-D=5xyz^2\)

\(\Leftrightarrow D=-xyz^2-5xyz^2\)

\(\Leftrightarrow D=-6xyz^2\)

a) \(2x^2+A=4,5x^2\)

\(\Rightarrow A=4,5x^2-2x^2\)

\(\Rightarrow A=2,5x^2\)

b) \(-2,5m^2n+B=5m^2n\)

\(\Rightarrow B=5m^2n-\left(-2,5m^2n\right)\)

\(\Rightarrow B=7,5m^2n\)

c) \(C-3,2ab^3=-11ab^3\)

\(\Rightarrow C=-11ab^3+3,2ab^3\)

\(\Rightarrow C=-7,8ab^3\)

d) \(-xyz^2-D=5xyz^2\)

\(\Rightarrow D=-xyz^2-5xyz^2\)

\(\Rightarrow D=-6xyz^2\)

#Ayumu

a: 2x^2+A=4,5x^2

=>A=4,5x^2-2x^2=2,5x^2

b: B-2,5m^2n=5m^2n

=>B=5m^2n+2,5m^2n=7,5m^2n

c: C-3,2ab^3=-11ab^3

=>C=-11ab^3+3,2ab^3=-7,8ab^3

d: -xyz^2-D=5xyz^2

=>D=-xyz^2-5xyz^2=-6xyz^2

5:

a: \(-120x^5y^4=20x^5y^2\cdot\left(-6y^2\right)\)

b: \(60x^6y^2=20x^5y^2\cdot3x\)

c: \(-5x^{15}y^3=20x^5y^2\cdot\left(-\dfrac{1}{4}x^{10}y\right)\)

d: \(2x^{12}y^{10}=20x^5y^2\cdot\left(\dfrac{1}{10}x^7y^8\right)\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

\(=>\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)

\(=>\left(3x+2\right)\left(x-4\right)=0\)

\(=>\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}3x=-2\\x=4\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=4\end{matrix}\right.\)

\(=>x\in\left\{\dfrac{-2}{3};4\right\}\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)(sửa đề)

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\)           \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Bài 1. (a) Điều kiện: \(x\ne\pm1\).

Ta có: \(A=\left(\dfrac{x-2}{x-1}-\dfrac{x+3}{x+1}+\dfrac{3}{x-1}\right):\left(1-\dfrac{x+3}{x+1}\right)\)

\(=\left(\dfrac{x-2+3}{x-1}-\dfrac{x+3}{x+1}\right):\dfrac{x+1-\left(x+3\right)}{x+1}\)

\(=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right):\dfrac{x+1-x-3}{x+1}\)

\(=\dfrac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{-2}{x+1}\)

\(=\dfrac{x^2+2x+1-x^2-2x+3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)

\(=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}=\dfrac{2}{1-x}\)

Vậy: \(A=\dfrac{2}{1-x}\)

(b) \(A=3\Leftrightarrow\dfrac{2}{1-x}=3\)

\(\Rightarrow1-x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\left(TM\right)\)

Vậy: \(x=\dfrac{1}{3}\)

Bài 2. (a) Phương trình tương đương với:

\(\dfrac{3\left(3x-2\right)}{12}+\dfrac{6\left(x+3\right)}{12}=\dfrac{4\left(x-1\right)}{12}+\dfrac{x+1}{12}\)

\(\Rightarrow3\left(3x-2\right)+6\left(x+3\right)=4\left(x-1\right)+x+1\)

\(\Leftrightarrow9x-6+6x+18=4x-4+x+1\)

\(\Leftrightarrow10x=-15\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\).

(b) Điều kiện: \(x\ne\pm1\). Phương trình tương đương với:

\(\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2+2}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow2\left(x+1\right)+2\left(x-1\right)=2x^2+2\)

\(\Leftrightarrow2x+2+2x-2=2x^2+2\)

\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)^2=0\Rightarrow x-1=0\Leftrightarrow x=1\left(KTM\right)\)

Vậy: Phương trình có tập nghiệm \(S=\varnothing\)

3:

1: =>15x-9x+6=45-10x+25

=>6x+6=-10x+70

=>16x=64

=>x=4

2: =>x^2+4x-16-16=0

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

3: ĐKXĐ: x<>4; x<>-4

\(PT\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)

=>x+4+x^2-2x-8=5x-4

=>x^2-x-4=5x-4

=>x^2-6x=0

=>x(x-6)=0

=>x=0 hoặc x=6

4: \(\Leftrightarrow5\left(4x+1\right)-x+2>=3\left(2x-3\right)\)

=>20x+5-x+2>=6x-9

=>19x+7>=6x-9

=>13x>=-16

=>x>=-16/13

2:

1: =7x(x-y)-5(x-y)

=(x-y)(7x-5)

2: =(x^2-y^2)-(4x-4y)

=(x-y)(x+y)-4(x-y)

=(x-y)(x+y-4)

3: =(x^2+2xy+y^2)-(2x+2y)+1

=(x+y)^2-2(x+y)+1

=(x+y-1)^2

\(2\left(x-1\right)-5\left(x+2\right)=-10\\ \Leftrightarrow2x-2-5x-10=-10\\ \Leftrightarrow-3x=2\\ \Leftrightarrow x=-\dfrac{2}{3}\)

A=-4x^2+4x-1+4

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

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

Dấu = xảy ra khi x=1/2

`a,`Số tiền mỗi bao gạo là :

`(x^{6}y^{5} - x^{5}y^{4}) : xy = x^{5}y^{4} - x^{4}y^{3}`

`b,`Khi `x=2;y=5` thì số tiền mỗi bao gạo :

`2^5 . 5^4 - 2^4 . 5^3 = 2^4 . 5^3 . (2 . 5 - 1)`

`=16 . 125 . 9`

`=18000`