phân tích đa thức thành nhân tử: (2-3a)^2-(3-a)^2
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
`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)`
\(a^3-b^3+3a^2+3ab+b^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+3\left(a^2+ab+b^2\right)\)
\(=\left(a-b+3\right)\left(a^2+ab+b^2\right)\)
Ta có:A= 3a+3b-a^2-ab
=>A= (3a-a^2)+(3b-ab)
=>A= a(3-a)+b(3-a)
=>A= (a+b)(3-a)
(2 - 3a)2 - (3 - a)2
= (2 - 3a - 3 + a)(2 - 3a + 3 - a)
= (-2a - 1)(1 - 4a)