Phân tích đa thức thành nhân tử
a) x^2-2019x-2020
b) y^3+y^2-10y+8
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
`#3107.101107`
`x^2 - y^2 + 10x - 10y`
`= (x^2 - y^2) + (10x - 10y)`
`= (x - y)(x + y) + 10(x - y)`
`= (x + y + 10)(x - y)`
_____
Sử dụng HĐT:
`A^2 - B^2 = (A - B)(A + B).`
\(a,x^2\left(x-2\right)-4x+8\\ =\left(x^2-4\right)\left(x-2\right)\\ =\left(x-2\right)^2\left(x+2\right)\\ b,x^2+7xy+10y^2\\ =x^2+2xy+5xy+10y^2\\ =x\left(x+2y\right)+5y\left(x+2y\right)\\ =\left(x+5y\right)\left(x+2y\right)\)
Bài 1:
a: \(5x^3+10xy=5x\left(x^2+2y\right)\)
b: \(x^2+14x+49-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+7+y\right)\left(x+7-y\right)\)
a) \(5x^2+5xy-x-y\)
\(=5x.\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
b) \(5x^2-10y+5y^2-20z^2\)
\(=5.\left(x^2-2y+y^2-4z^2\right)\)
Đề sai ở đâu đó.
c) \(4x^2-y^2+4x+1\)
\(=\left(4x+4x^2+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+y+1\right)\left(2x-y+1\right)\)
a: x^2+4xy-21y^2
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b: \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
c: \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
=x(x+5y)-3y(x+5y)
=(x+5y)(x-3y)
d: \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
=x(x-2y)-5y(x-2y)
=(x-2y)(x-5y)
a) \(x^2+4xy-21y^2\)
\(=x^2+7xy-3xy-21y^2\)
\(=x\left(x+7y\right)-3y\left(x+7y\right)\)
\(=\left(x+7y\right)\left(x-3y\right)\)
b) \(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(5x+y\right)\left(x+y\right)\)
c) \(x^2+2xy-15y^2\)
\(=x^2+5xy-3xy-15y^2\)
\(=x\left(x+5y\right)-3y\left(x+5y\right)\)
\(=\left(x+5y\right)\left(x-3y\right)\)
d) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^2-2019x-2020\)
\(=x^2+x-2020x-2020\)
\(=x\left(x+1\right)-2020\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2020\right)\)
Bạn hộ mình câu b) được không ^^