a)\(x^2+7x+12\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

a) x² + 7x + 12 = x² + 3x + 4x + 12

= (x² + 3x) + (4x + 12)

= x(x + 3) + 4(x + 3)

= (x + 3)(x + 4)

vậy.....

b) \(3x^2+2x-5=3\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)

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

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e) \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

b: \(3x^2+2x-5\)

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

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

c: \(3-2x-x^2\)

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

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

d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)

\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)

Đặt:   \(x^2+5x+5=a\)Khi đó ta có:

\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)

tự thay trở lại

Mình sửa: Bài 1
2)x²+3x-15

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

\(x^2-7x+10\)

\(=x^2-2x-5x+10\)

\(=x\left(x-2\right)-5\left(x-2\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

học tốt

1) x² -7x + 10 = x² - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

2) x² + 3x + 2 = x² + 2x + x  + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)

3) x² - 7x + 12 = x² - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)

4) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

5) 16x - 5x² - 3 = 15x - 5x² + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

6) 6x² + 7x - 3 = 6x² - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)  

7) 3x² - 3x - 6 = 3x² - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)

8) 3x² + 3x - 6 = 3x² - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)

9) 6x² - 13x + 6 = 6x² - 9x -  4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3) 

10) 6x² + 15x  + 6 = 6x² + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)

11) 6x² - 20x + 6 = 6x² - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)

12) 8x² + 5x - 3 = 8x² + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)

a) đề thế này\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+11=t\)vào (1) ta được:

\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

b) Phân tích sẵn rồi còn phân tích gì nưa=))

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)( Làm đề theo Lê Tài Bảo Châu )

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left[\left(x^2+7x+11\right)-1\right]\left[\left(x^2+7x+11\right)+1\right]-24\)

\(=\left(x^2+7x+11\right)^2-1-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

f)\(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)

i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)

h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)

g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

f)\(x^2-5x-14=x^2-7x+2x-14\)

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

i)\(x^2-7x+10=x^2-5x-2x+10\)

                              \(=\left(x-2\right)\left(x-5\right)\)

h)\(x^2-7x+12=x^2-4x-3x+12\)

                              \(=\left(x-3\right)\left(x-4\right)\)

g)\(x^2+6x+5=x^2+x+5x+5\)

                           \(=\left(x+5\right)\left(x+1\right)\)

f) \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

\(=\left(x^2-7x\right)+\left(2x-14\right)\)

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

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

i) \(x^2-7x+10\)

\(=x^2-5x-2x+10\)

\(=\left(x^2-5x\right)-\left(2x-10\right)\)

\(=x\left(x-5\right)-2\left(x-5\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

h) \(x^2-7x+12\)

\(=x^2-3x-4x+12\)

\(=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-4\right)\left(x-3\right)\)

g) \(x^2+6x+5\)

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

\(=\left(x^2+x\right)+\left(5x+5\right)\)

\(=x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+5\right)\left(x+1\right)\)

a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)

b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)

\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)

a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)

b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)

=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)