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 \(n\).

a) \(\left(x+2\right)^3-x^2.\left(x+6\right)\)

\(=x^3+6x^2+12x+8-x^3-6x^2\)

\(=12x+8\)

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

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

\(=-3x^3+2x^2-5x-5\)

a,  xét tam giác FIC và tam giác EBC có : ^ECB chung

^FIC = ^EBC = 90

=> tam giác FIC đồng dạng với tam giác EBC (g-g)

b, có ^IFC + ^ICF = 90

^ICD + ^ICF = 90

=> ^IFC = ^ICD

xét tam giác FIC và tg CID có : ^FIC = ^CID = 90

=> tam giác FIC đồng dạng với tg CID (g-g)

=> IC/IF = ID/IC

=> IC^2 = IF.ID