phân tích các đa thức sau thành nhân tử: x^2 +2021x-2022
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x4+2012x2+2012x+2012
=(x4-x)+(2012x2+2012x+2012)
=x(x3-1)+2012(x2+x+1)
=x(x-1) (x2+x+1) + 2012 (x2+x+1)
=(x2+x+1) [x(x-1)+2012]
=(x2+x+1) (x2-x+2012)
x4 + 2021x2 - 2020x + 2021
= (x4 + x) + 2021(x2 - x + 1)
= x(x3 + 1) + 2021(x2 - x + 1)
= x(x + 1)(x2 - x + 1) + 2021(x2 - x + 1)
= (x2 + x + 2021)(x2 - x + 1)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
Ta có: \(x^2-3x-28\)
\(=x^2-7x+4x-28\)
\(=x\left(x-7\right)+4\left(x-7\right)\)
\(=\left(x-7\right)\left(x+4\right)\)
x^2- 4y^2 + 4xy
= x^2 + 4xy - 4y^2
=x^2 + 2x2y - (2y)^2
= ( x - 2y )^2
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=\left(x-2\right)\left(8x-1\right)\)
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)+5x\left(x-2\right)=\left(x-2\right)\left(3x=1+5x\right)=\left(x-2\right)\left(8x-1\right)\)
\(=x^2-x+2022x-2022\\ =x\left(x-1\right)+2022\left(x-1\right)\\ =\left(x+2022\right)\left(x-1\right)\)