Chia
( x 2-y 2+6y-9):(x-y+3)
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\(\left(x^2-y^2+6y-9\right):\left(x-y+3\right)\\ \\=\left[x^2-\left(y^2-6y+9\right)\right]:\left(x-y+3\right)\\ \\=\left[x^2-\left(y-3\right)^2\right]:\left(x-y+3\right)\\ \\=\left(x-y+3\right)\left(x+y-3\right):\left(x-y+3\right)\\ \\=x+y-3\)
(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
(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
\(\left(x^2-y^2+6y-9\right):\left(x-y+3\right)\)
\(=\left[x^2-\left(y^2-6y+9\right)\right]:\left(x-y+3\right)\)
\(=\left[x^2-\left(y-3\right)^2\right]:\left(x-y+3\right)\)
\(=\left(x+y-3\right)\left(x-y+3\right):\left(x-y+3\right)\)
=\(x+y-3\)
TL:
1)\(\left(y^2-6y+9\right)-\left(3-y\right)^2=\left(y-3\right)^2-\left(3-y\right)^2\)
\(=\left(y-3+3-y\right)\left(y-3-3+y\right)=0.\left(2y-6\right)=0\)
2)\(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=\left(x-3\right)^2-x^2+16\)
\(=\left(x-3+x\right)\left(x-3-x\right)+16=\left(2x-3\right).\left(-3\right)+16=-6x+9+16\)
\(=-6x+25\)
hc tốt
\(1,\left(y^2-6x+9\right)-\left(3-y\right)^2\)
\(=\left(y-3\right)^2-\left(y-3\right)^2=0\)
\(2,\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16=-6x+21\)
\(3...\)\(< ->1\)
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
\(x^3-7x-6=0\)
\(x^3-3x^2+3x^2+2x-9x-6=0\)
\(x^2.\left(x-3\right)+3x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x+3\right).\left(x^2+3x+2\right)=0\Rightarrow\left(x-3\right).\left(x^2+3x+x+2\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x+1\right).\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\text{hoặc }x=-2\)
(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)(x - y + 3):(x - y + 3)
= x + y - 3
\(\left(x^2-y^2+6xy-9\right):\left(x-y+3\right)=\left[\left(x-y\right)^2-9\right]:\left(x-y+3\right)\)
\(=\left(x-y+3\right)\left(x-y-3\right):\left(x-y+3\right)=x-y-3\)