x^2-xy-2x+2y
x^2-9y^2+6y-1
Phân tích đa thức thành nhân tử
x^2-2xy+y^2-2x+2y
x^2-4x+4-x^2y+2xy
ax^2-3axy-x^2+6xy-9y^2
2a^2x-5a^2y-4x^2+30xy-25y^2
a) Ta có: \(x^2-2xy+y^2-2x+2y\)
\(=\left(x-y\right)^2-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2-4x+4-x^2y+2xy\)
\(=\left(x-2\right)^2-xy\left(x-2\right)\)
\(=\left(x-2\right)\left(x-2-xy\right)\)
c) Ta có: \(ax^2-3axy-x^2+6xy-9y^2\)
\(=ax\left(x-3y\right)-\left(x^2-6xy+9y^2\right)\)
\(=ax\left(x-3y\right)-\left(x-3y\right)^2\)
\(=\left(x-3y\right)\left(ax-x+3y\right)\)
d) Ta có: \(2a^2x-5a^2y-4x^2+30xy-25y^2\)
\(=a^2\left(2x-5y\right)-\left(4x^2-30xy+25y^2\right)\)
\(=a^2\left(2x-5y\right)-\left(2x-5y\right)^2\)
\(=\left(2x-5y\right)\left(a^2-2x+5y\right)\)
Rút gọn phân thức \(\dfrac{x^2-9y^2}{x^2+xy-6y^2}\)
\(\dfrac{x^2-9y^2}{x^2+xy-6y^2}=\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-2y\right)\left(x+3y\right)}=\dfrac{x-3y}{x-2y}\)
1) x^2-4xy-x+3y^2+3y
2) 6x^2+xy -7x-2y^2+7y-5
3) 2a^2+5ab-3b^2-7b-2
4) 6x^2-xy-2y^2+3x-2y
5) 2x^2 - 3xy-4x-9y^2-6y
Giúp mk với mk đang cần gấp
1: \(=\left(x-3y\right)\left(x-y\right)-\left(x-3y\right)=\left(x-3y\right)\left(x-y-1\right)\)
4: \(=6x^2-4xy+3xy-2y^2+3x-2y\)
\(=\left(3x-2y\right)\left(2x+y\right)+3x-2y=\left(3x-2y\right)\left(2x+y+1\right)\)
Bài 1 :Thực hiện phép tính :
a)(2x-1)^2(2x+3)(2x-5)
b)(x+2)^3-x(x-1)(x+1)
Bài 2:Phân tích đa thức thành nhân tử :
a)x^3-4x
b)x^2-xy-6x+6y
c)x^2-6x-y^2+9
d)x^2-y^2+2x-2y
e)25x^2-9y^2-10x-1
g)x^3-xy^2+6xy-9x
i)x^2-4x-5
h)2x^2-7x+3
x2 - xy - 6y2 / x2 - 9y2
\(\frac{\text{x2-xy-6y2}}{\text{x2-9y2 }}=\frac{\text{x(x-2y)+3y(x-2y)}}{\text{x2-(3y)2 }}=\frac{\text{(x+3y)(x-2y)}}{\text{(x-3y)(x+3y)}}=\frac{(x-2y)}{\left(x-3y\right)}\)
rút gọn phân thức \(\dfrac{x^2-xy-6y^2}{x^2-9y^2}\) thu được kết quả là
\(\dfrac{x^2-xy-6y^2}{x^2-9y^2}=\dfrac{\left(x-3y\right)\left(x+2y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x+2y}{x+3y}\)
thực hiện phép tính
a)\(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
b)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}\)
c) \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}\)
a/ \(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=-\dfrac{4x+2}{x\left(1-3x\right)}\cdot\dfrac{1-3x}{x^2+3x}=-\dfrac{4x^2+2}{x\left(x^2+3x\right)}\)
b/ \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}=-\dfrac{2\left(2x+3y\right)}{1-x}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(2x+3y\right)^2}=\dfrac{-2\left(x+1\right)}{2x+3y}=\dfrac{-2x-2}{2x+3y}\)
c/ \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}=\dfrac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}\cdot\dfrac{2x+y}{x\left(x^2+xy+y^2\right)}=\dfrac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y}\cdot\dfrac{1}{x\left(x^2+xy+y^2\right)}=\dfrac{x-y}{y}\)
- ( 2x + 1 ) (3y - 2 ) + 6y (x -1 ) = 15 - ( 1+9y)
\(-\left(2x+1\right)\left(3y-2\right)+6y\left(x-1\right)=15-\left(1+9y\right)\)
\(\Leftrightarrow-\left(6xy-4x+3y-2\right)+6xy-6y=15-1-9y\)
\(\Leftrightarrow-6xy+4x-3y+2+6xy-6y-14+9y=0\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
Chúc bạn học tốt!!!
Tk cho mk nha... Cảm ơn nhìu...
bài 2; phân tích đa thức sau thành nhân tử
a, x mũ 2 + 7x + 7y - y mũ 2
b, x mũ 2 - 2x - 9y mũ 2 + 6y
c, x mũ 2 - xy +x mũ 3 - 3x mũ 2 y + 3x mũ 2 y - y mũ 3
\(a,x^2+7x+7y-y^2\)
\(=x^2-y^2+7\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
\(b,x^2-2x-9y^2+6y\)
\(=x^2-\left(3y\right)^2-2\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-2\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-2\right)\)
\(c,x^2-xy+x^3-3x^{2y}+3x^{2y}-y^3\)
\(=x\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x+x^2+xy+y^2\right)\)