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 \(N=y^2-x^2\)

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

a)  \(A=x^2+2xy+y^2=\left(x+y\right)^2=\left(-1\right)^2=1\)

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

c)  \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

a) ĐKXĐ: \(x\ne0\)

 \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

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

\(=\dfrac{10x-1}{6x}\)

b) ĐKXĐ: \(x,y\ne0\)

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

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

\(=\dfrac{x-y}{2xy}\)

a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

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

\(=\dfrac{8x+2+2x-3}{6x}\)

\(=\dfrac{10x-1}{6x}\)

b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

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

\(=\dfrac{x-y}{2xy}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a: \(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{4}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)

\(=-\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{-1}{18}\)

\(=\dfrac{-1}{72}\)

b: \(B=\left(-1\right)^2\cdot3^2+\left(-1\right)\cdot3+\left(-1\right)^3+3^3\)

\(=9-3-1+27=36-4=32\)

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 )\(2x\left(xy-3\right)+3xy\left(x+1-y\right)+3x\left(y^2-1\right)=2x^2y-6x+3x^2y+3xy-3xy^2+3xy^2-3x=5x^2y-9x+3xy\)

=> Phụ thuộc vào giá trị của biến

b) \(\left(x+2y\right)\left(x-2y\right)-x\left(x+4y^2\right)+5=x^2-4y^2-x^2-4xy^2+5=-4y^2-4xy^2+5\)

=> Phụ thuộc vào giá trị của biến

c) \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)=27x^3+8-9x^2+4=27x^3-9x^2+12\)

=> Phụ thuộc vào giá trị của biến

a: Ta có: \(2x\left(xy-3\right)+3xy\left(x-y+1\right)+3x\left(y^2-1\right)\)

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

\(=5x^2y+3xy-9x\)

c: Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(3x+2\right)\)

\(=27x^3+8-9x^2+4\)

\(=27x^3-9x^2+12\)