Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

a) 5xy ( x - y ) - 2x + 2y

= 5xy ( x - y ) - 2 ( x - y )

= ( x - y ) ( 5xy - 2 )

b) 6x-2y-x(y-3x)

= 2 ( y - 3x ) - x ( y - 3x )

= ( y - 3x ( ( 2 - x )

c)  x² + 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

= ( x - 2 ) ( 3y + z )

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

(2-3x)(2+3x)=0

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

x(x+1/3)(x-1/3)=0

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

đề dài nên T giải câu a thôi bn tự làm tiếp mấy câu khác nhé 

2x^2 - 2y^2 - 6x - 6y 

= 2(x^2-y^2) - 6(x+ y) 

= 2(x-y)(x+y) - 6(x+y) 

= (2(x-y)-6) (x+y)

ko biết

Bài 10 :

Câu a :

\(5xy\left(x-y\right)-2x+2y\)

\(=5xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(5xy-2\right)\)

Câu b :

\(6x-2y-x\left(y-3x\right)\)

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

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

Câu c :

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

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

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

Câu d :

\(3xy+2z-6y-xz\)

\(=\left(3xy-6y\right)-\left(xz-2z\right)\)

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

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

Bài 11 :

Câu a :

\(4-9x^2=0\)

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

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

Vậy ........................

Câu b :

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{2}=0\)

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

Vậy........................

Câu c :

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..................

Câu d :

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

\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy................................

Câu e :

\(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy........................

Câu f :

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

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

\(\Leftrightarrow2x\left(4x-2y\right)=0\)

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

Vậy..........................

P(x)+Q(x)

=3x^2y-2x+5xy^2-7y^2+3xy^2-7y^2-9x^2y-x-5

=8xy^2-14y^2-6x^2y-3x-5

=>Chọn A

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)