Phân tích đa thức thành nhân tử :
x(x-2021)-x+2021
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(5x^3+10xy=5x\left(x^2+2y\right)\)
b: \(x^2+14x+49-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+7+y\right)\left(x+7-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5x\left(x-2021\right)-x+2021=0\)
\(5x\left(x-2021\right)-\left(x-2021\right)=0\)
\(\left(x-2021\right)\left(5x-1\right)=0\)
\(\orbr{\begin{cases}x-2021=0\\5x-1=0\end{cases}\orbr{\begin{cases}x=2021\left(TM\right)\\x=\frac{1}{5}\left(TM\right)\end{cases}}}\)
Trả lời:
\(5x\left(x-2021\right)-x+2021=0\)
\(\Leftrightarrow5x\left(x-2021\right)-\left(x-2021\right)=0\)
\(\Leftrightarrow\left(x-2021\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2021=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2021\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = 2021; x = 1/5 là nghiệm của pt.
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+a+1=0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}=0\Rightarrow a\in C\)
Vì vậy P không tồn tại
Lớp 8 nên làm như này nhé :))
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^3-2x^2+2x-1\right)\left(x-1\right)\)
\(=\left(x^3-x^2-x^2+x+x-1\right)\left(x-1\right)\)
\(=\left(x^2-x+1\right)\left(x-1\right)^2\)
c)
\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)
\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)
\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)
\(=\left(6x^2-x-1\right)\left(x+3\right)\left(x-2\right)\)
\(=\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\left(x-2\right)\)
b)
\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)
\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)
\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)
CÓ: \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)
=> \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)
=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)
2) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)
\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) = \(x^2-1=\left(x-1\right)\left(x+1\right)\)
2) \(=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3)
\(=x^4-x+x^2+x+1=x\left(x^3-1\right)+x^2+x+1=x\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(=x^5-x^2+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
1.\(x^2-2021+2020=x^2-1=\left(x+1\right)\left(x-1\right)\)
2. \(x^4+64=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3. \(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2+x+1\right)\)
4. \(x^5+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
=x(x-2021)-(x-2021)
=(x-2021)(x-1)