phân tích đa thức sau thành nhân tử 1 x^2 (4-x) + 9 (4-x)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-9-4\left(x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3-4\right)\)
\(=\left(x+3\right)\left(x-7\right)\)
x2 - 9 - 4(x + 3)
= (x2 - 9) - 4(x + 3)
= (x2 - 32) - 4(x + 3)
= (x - 3) (x + 3) - 4(x + 3)
= (x + 3) (x - 3 - 4)
= (x + 3) ( x - 7)
ko bít đúng ko !! 547467474856856856856835675743634634645765876976946426436
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-9-4\left(x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3-4\right)\)
\(=\left(x+3\right)\left(x-7\right)\)
\(x^2-9-4\left(x+3\right)\)
\(=\left(x^2-9\right)-4\left(x+3\right)\)
\(=\left(x^2-3^2\right)-4\left(x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3-4\right)\)
\(=\left(x+3\right)\left(x-7\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên viết lại đa thức bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
\(x^2\left(4-x\right)+9\left(4-x\right)=\left(x^2+9\right)\left(4-x\right)\)