Làm tính chia:
c,( x^2 - y^2 + 6y - 9) : ( x-y + 3)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(x2 -y2 +6y -9) : (x-y+3)
= [x2-(y2-6y+9)] : (x-y+3)
=[x2-(y-3)2 ] : (x-y+3)
=[(x-y+3)(x+y-3)] :(x-y+3)
=x+y-3
\(\left(x^2-y^2+6xy-9\right):\left(x-y+3\right)=\left[\left(x-y\right)^2-9\right]:\left(x-y+3\right)\)
\(=\left(x-y+3\right)\left(x-y-3\right):\left(x-y+3\right)=x-y-3\)
\(x^3-7x-6=0\)
\(x^3-3x^2+3x^2+2x-9x-6=0\)
\(x^2.\left(x-3\right)+3x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x+3\right).\left(x^2+3x+2\right)=0\Rightarrow\left(x-3\right).\left(x^2+3x+x+2\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x+1\right).\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\text{hoặc }x=-2\)
Bài 1:
a: =>x^3-x-6x-6=0
=>x(x-1)(x+1)-6(x+1)=0
=>(x+1)(x-3)(x+2)=0
hay \(x\in\left\{-1;3;-2\right\}\)
b: \(\Leftrightarrow x^2-6x+9+y^2+6y+9=0\)
=>(x-3)^2+(y+3)^2=0
=>x=3 và y=-3
(x2 - y2 + 6y - 9) : (x - y + 3)
= [x2 - (y2 - 6y + 9)] : (x - y + 3)
= [x2 - (y - 3)2] : (x - y + 3)
= (x - y + 3)(x + y - 3) : (x - y + 3)
= x + y - 3
\(\left(x^2-y^2+6y-9\right):\left(x-y+3\right)\)
\(=\left[x^2-\left(y^2-6y+9\right)\right]:\left(x-y+3\right)\)
\(=\left[x^2-\left(y-3\right)^2\right]:\left(x-y+3\right)\)
\(=\left(x+y-3\right)\left(x-y+3\right):\left(x-y+3\right)\)
=\(x+y-3\)
ban chu y nhe
x^2-y^2+6x+9 = (x^2 +6x+9)-y^2(sử dụng hđt a^2+2ab+b^2=(a+b)^2 )
=(x+3)^2-y^2 (sử dụng hđt a^2 - b^2= (a+b)(a-b)
=(x+3+y)(x+3-y)
=>(x^2 -y^2+6x+9)/(x+y+3)=x-y+3
chúc may mắn
mik ko biết viết số mũ thông cảm
\(\left(x^2-y^2+6y-9\right):\left(x-y+3\right)\\ \\=\left[x^2-\left(y^2-6y+9\right)\right]:\left(x-y+3\right)\\ \\=\left[x^2-\left(y-3\right)^2\right]:\left(x-y+3\right)\\ \\=\left(x-y+3\right)\left(x+y-3\right):\left(x-y+3\right)\\ \\=x+y-3\)