Bài 1:

a: \(5x^3+10xy=5x\left(x^2+2y\right)\)

b: \(x^2+14x+49-y^2\)

\(=\left(x+7\right)^2-y^2\)

\(=\left(x+7+y\right)\left(x+7-y\right)\)

\(5x\left(x-2021\right)-x+2021=0\)

\(5x\left(x-2021\right)-\left(x-2021\right)=0\)

\(\left(x-2021\right)\left(5x-1\right)=0\)

\(\orbr{\begin{cases}x-2021=0\\5x-1=0\end{cases}\orbr{\begin{cases}x=2021\left(TM\right)\\x=\frac{1}{5}\left(TM\right)\end{cases}}}\)

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.

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)

a, \(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)

b, =  \(y^2-4x^2+4x^2=y^2\)

Thay y = 10 vào BT trên, ta có:

\(y^2=10^2=100\)

Vậy giá trị của BT là 100

\(a^2+a+1=0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}=0\Rightarrow a\in C\)

Vì vậy P không tồn tại

Lớp 8 nên làm như này nhé :))

2021ⁿ⁺¹-2021ⁿ

=2021ⁿ.2021-2021ⁿ 

=2021ⁿ.(2021-1)

=2021ⁿ.2020\(⋮\)2020 với mọi n thuộc N

 Vậy 2021ⁿ⁺¹-2021ⁿ chia hết cho 2000 , với mọi n thuộc N

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.

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) \(A=x^2+2y^2+2xy-2y+2021=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2020=\left(x+y\right)^2+\left(y-1\right)^2+2020\ge2020\)

\(minA=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=2.\left(x+4\right).4\left(x-1\right)=8\left(x-1\right)\left(x+4\right)\)

Bài 1: 

Ta có: \(\left(3x+2\right)^2-\left(x-6\right)^2\)

\(=\left(3x+2-x+6\right)\left(3x+2+x-6\right)\)

\(=\left(2x+8\right)\left(4x-4\right)\)

\(=8\left(x+4\right)\left(x-1\right)\)

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. \(x^5+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)