3x³ + 6x² + 3x - 12xy² 

= 3x(x² + 2x + 1 - 4y²) 

= 3x[(x + 1)² - (2y)²]

= 3x(x + 1 + 2y)(x - 2y + 1)

\(3x^3+6x^2+3x-12xy^2\)

\(=3x\left(x^2+2x+1-4y^2\right)\)

\(=3x\left[\left(x+1\right)^2-\left(2y\right)^2\right]\)

\(=3x\left(x+1-2y\right)\left(x+1+2y\right)\)

\(2\left(x+1\right)\left(1-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\1-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy \(S=\left\{-1;1\right\}\)

giải

x+1=0

1-x=0

vây ta có -1và1

x⁵ + 27x² 

 = x²(x³ + 27) 

 = x²(x + 3)(x² - 3x + 9) 

Bài 2. 

\(n^4-2n^3-n^2+2n=n\left(n^3-2n^2-n+2\right)=n\left[n^2\left(n-2\right)-\left(n-2\right)\right]\)

\(=n\left(n-2\right)\left(n^2-1\right)=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

là tích của \(4\)số nguyên liên tiếp nên trong đó có ít nhất \(1\)thừa số chia hết cho \(4\), \(1\)thừa số chia hết cho \(3\), \(1\)thừa số chia hết cho \(2\)nhưng không chia hết cho \(4\)

do đó \(A\)chia hết cho \(2.3.4=24\).

Ta có đpcm. 

Bài 1: 

\(2-x=2\left(x-2\right)^3\)

\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2\left(x-2\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\pm\sqrt{\frac{1}{2}}+2\end{cases}}\)

24x³ - 3

= 24x³ - 12x² + 12x² - 6x + 6x -3 

= 12x² (2x - 1) + 6x(2x -1) + 3(2x-1)

=(2x-1)(12x² + 6x + 3)

= 3(2x-1)(4x² + 2x + 1) 

11, \(a^4-16b^4=\left(a^2+4b^2\right)\left(a^2-4b^2\right)=\left(a^2+4b^2\right)\left(a-2b\right)\left(a+2b\right)\)

12, \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

13, \(\left(a-2b\right)^2-4b^2=\left(a-2b-2b\right)\left(a-2b+2b\right)=\left(a-4b\right).a\)

14, \(\left(a+3b\right)^2-9b^2=\left(a+3b-3b\right)\left(a+3b+3b\right)=a.\left(a+6b\right)\)

15, \(\left(a-5b\right)^2-16b^2=\left(a-5b-4b\right)\left(a-5b+4b\right)=\left(a-9b\right)\left(a-1\right)\)

16, \(36a^2-\left(3a-2b\right)^2=\left(6a-3a+2b\right)\left(6a+3a-2b\right)=\left(3a+2b\right)\left(9a-2b\right)\)

17, \(4a^2-\left(a+b\right)^2=\left(2a-a-b\right)\left(2a+a+b\right)=\left(a-b\right)\left(3a+b\right)\)

18, \(49a^2-\left(2a-b\right)^2=\left(7a-2a+b\right)\left(7a+2a-b\right)=\left(5a+b\right)\left(9a-b\right)\)

Trả lời:

\(A=2x^2+x=2\left(x^2+\frac{1}{2}x\right)=2\left(x^2+2.x\frac{1}{4}+\frac{1}{16}-\frac{1}{16}\right)\)

\(=2\left[\left(x+\frac{1}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy \(Min_A=-\frac{1}{8}\Leftrightarrow x=-\frac{1}{4}\)

Đề bài có chút vấn đề bạn nhé. 

\(a+b+c=0\Leftrightarrow c=-\left(a+b\right)\)

\(a^4+b^4+c^4=a^4+b^4+\left(a+b\right)^4\)

\(=2\left(a^4+b^4+2a^3b+2ab^3+3a^2b^2\right)\)

\(=2\left(a^2+b^2+ab\right)^2\)

Do đó \(\frac{a^4+b^4+c^4}{2}\)là số chính phương. 

a) \(x^2-6x-y^2-4y+5=x^2-6x+9-\left(y^2+4x+4\right)=\left(x-3\right)^2-\left(y+2\right)^2\)

b) \(4a^2-12a-b^2+2b+8=4a^2-12a+9-\left(b^2-2b+1\right)=\left(2a-3\right)^2-\left(b-1\right)^2\)

\(Q=\left(2n-1\right)\left(2n+3\right)-\left(4n-5\right)\left(n+1\right)+3\)

\(=4n^2+4n-3-\left(4n^2-n-5\right)+3\)

\(=5n+5⋮5\)với mọi số nguyên \(n\).