4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

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

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

1. = 2(a+b)

2. =2(a-b)

3.=2(a+2b-3c)

4.=3(a-2b-3c)

5.=-4(a+2b+3c)

6.=-5(x+2xy+3y)

7.=-7(a+2ab+3b)

8.=6(xy-2x-3y)

9.=8(xy-3y+2x)

10.=9(ab-2a+1)

Lần sau bn cần ghi đề rõ ràng hơn

11.=x(y-1)

12.=a(x+1)

13.=m(x+y+1)

14.=-a(x+y+1)

15.=-a(x²+x+1)

16.=-2a(x+2y)

17.=2a(x-y+1)

18.=2(2ax-ay-2)

19.=5a(1-2x-3)

20.=-2ab(a+2b+3)

Bài 1:

\(\frac{15ab+5b^2}{9a^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a\right)^2-b^2}=\frac{5b\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{5b}{3a-b}\)

\(\frac{3x^2-3y^2}{9x+9y}=\frac{3\left(x^2-y^2\right)}{9\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{3\left(x+y\right)}=\frac{x-y}{3}\)

\(\frac{m^2-4m+4}{2x-4}=\frac{\left(x-2\right)^2}{2\left(x-2\right)}=\frac{x-2}{2}\)

a: \(12x^4y^3+12x^3y^3+3x^2y^3\)

\(=3x^2y^3\cdot4x^2+3x^2y^3\cdot4x+3x^2y^3\cdot1\)

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

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

b: \(x^4+xy^3-x^3y-y^4\)

\(=\left(x^4+xy^3\right)-\left(x^3y+y^4\right)\)

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

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

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

Phân tích đa thức thành nhân tử

a. 3ab ( x+ y) - 6ab ( y+ x)

   =( x + y) ( 3ab - 6ab )

     = ( x +y ) ( - 3ab)

b.7a (x - 3)+a²(x² - 9)

  =7a( x- 3) + a² ( x² - 3²)

  =7a ( x - 3 ) + a² ( x- 3 ) ( x+3 )

   = ( x- 3) . 7a + a² ( x + 3)

    = ( x- 3) ( 7a +a²x + 3a²)

c. 34 (x + y) -x -y

  = 34 ( x+ y) - ( x+y)

 =(x +y ) ( 34 - 1) = 33 ( x+ y)

  d. 25 x⁴  - 94²

     =( 5x² )² - 94²

     =( 5x² - 94 ) ( 5x²+94)

   e.( 5a - b )² - ( 2a +3b)²

      =( 5a -b -2a - 3b) (5a -b + 2a + 3b)

       =(3a - 4b) (7a+ 2b)

 k. 2² -3a - b² +3b

    =( 2² - b² ) + ( -3a +3b)

    =( 2-b) (2+b) + 3( -a +b)

mk làm đầu tiên nhớ tick cho mk nhé!!

dài quá, làm từ từ nhé

1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)

\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)

\(=4ab\left(a-2b\right)\)

2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)

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

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

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

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

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

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

4, câu này đề thiếu

5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)

\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)

\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)

\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)

\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)

6, \(\left(x+y-2z\right)^2+\left(x+y+2z\right)^2-16z^2\)

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

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

\(=\left(x+y-2z\right)\left(x+y-2z+x+y+6z\right)\)

\(=\left(x+y-2z\right)\left(2x+2y+4z\right)\)

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

7,\(=a^2x^2+6axy+9y^2-\left(-6ax^2-6ay^2+x^2+y^2\right)+9x^2-6axy+a^2y^2\)

\(=a^2x^2+6axy+9y^2+6ax^2+6ay^2-x^2-y^2+9x^2-6axy+a^2y^2\)

\(=a^2x^2+6ax^2+8x^2+a^2y^2+6ay^2+8y^2\)\(=x^2\left(a^2+6a+8\right)+y^2\left(a^2+6a+8\right)\)

\(=\left(x^2+y^2\right)\left(a^2+6a+8\right)\)\(=\left(x^2+y^2\right)\left(a^2+2a+4a+8\right)\)

\(=\left(x^2+y^2\right)\left[a\left(a+2\right)+4\left(a+2\right)\right]=\left(x^2+y^2\right)\left(a+2\right)\left(a+4\right)\)

1.

\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

\(\Rightarrow\dfrac{1}{2a+c}=\dfrac{1}{2b}=\dfrac{1}{b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)

Bài 2 đề sai

Ở phân thức thứ 2 không thể là \(\dfrac{y+3x-x}{x}\)

Bài 2:

\(P=\dfrac{x+3y}{y}\cdot\dfrac{y+3z}{z}\cdot\dfrac{z+3x}{x}=\dfrac{\left(x+3y\right)\left(y+3z\right)\left(z+3x\right)}{xyz}\)

Với \(x+y+z=0\)

\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}\\ \Leftrightarrow\dfrac{x+3y+x+y}{z}=\dfrac{y+3z+y+z}{x}=\dfrac{z+3x+x+z}{y}\\ \Leftrightarrow\dfrac{2\left(x+2y\right)}{z}=\dfrac{2\left(y+2z\right)}{x}=\dfrac{2\left(z+2x\right)}{y}\\ \Leftrightarrow\dfrac{2\left(y-z\right)}{z}=\dfrac{2\left(z-x\right)}{x}=\dfrac{2\left(x-y\right)}{y}\\ \Leftrightarrow\dfrac{2y-2z}{z}=\dfrac{2z-2x}{x}=\dfrac{2x-2y}{y}\\ \Leftrightarrow\dfrac{2y}{z}-2=\dfrac{2z}{x}-2=\dfrac{2x}{y}-2\\ \Leftrightarrow\dfrac{2y}{z}=\dfrac{2z}{x}=\dfrac{2x}{y}\\ \Leftrightarrow\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}\Leftrightarrow x=y=z=0\left(\text{trái với GT}\right)\)

Với \(x+y+z\ne0\)

\(\Leftrightarrow\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y-z=3z\\y+3z-x=3x\\z+3x-y=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=4z\\y+3z=4x\\z+3x=4y\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{4x\cdot4y\cdot4z}{xyz}=64\)