ban nao giup minh vs mjnh vs

1. a) 7x² - 5x - 2 = 7x² - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)

2. 5(2x - 1)² - 3(2x - 1) = 0

<=> (2x - 1).[5(2x - 1) - 3] = 0

<=> (2x - 1).(10x - 8) = 0

<=> (2x - 1) = 0 hoặc (10x - 8) = 0

<=> x = 1/2 hoặc x = 4/5

3. x² - 4x + 7 = (x² - 4x + 4) + 3 = (x - 2)² + 3

Do: (x - 2)² > hoặc = 0 (với mọi x)

Nên (x - 2)² + 3 > hoặc = 3 (với mọi x)

Hay (x - 2)² + 3 > 0 (với mọi x)  => đpcm

\(7x^2-5x-2\)

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

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

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

\(a^2-6a+5=\left(a^2-5a\right)-\left(a-5\right)=a\left(a-5\right)-\left(a-5\right)=\left(a-1\right)\left(a-5\right)\) 

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

\(4a^2+4a-3=4a^2-2a+\left(6a-3\right)=2a\left(2a-1\right)+3\left(2a-1\right)=\left(2a+3\right)\left(2a-1\right)\)

X² - 6x + 5

= x² - 6x + 5 + 4 - 4 

= x² - 6x + 9 - 2²

= ( x - 3 )² - 2² 

= ( x - 3 - 2 ) ( x - 3 + 2 ) 

X² - 7x + 12

= X² - 7x + 12 + 0.25 - 0,25

= X² - 7x + 12,25 - ( 0,5 )²

= ( X - 3,5 )² -  ( 0,5 )²

= ( X - 3,5 - 0,5 ) ( X - 3,5 + 0,5 )

vì để dễ tính hơn nha

((: Dễ tính hơn ấy ạ:")?

a) \(x^3+4x^2-21x\)

\(=x\left(x^2+4x-21\right)\)

\(=x\left(x^2-3x+7x-21\right)\)

\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)

\(=x\left(x-3\right)\left(x+7\right)\)

b) \(5x^3+6x^2+x\)

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

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

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

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

c) \(x^3-7x+6\)

\(=x^3+2x^2-3x-2x^2-4x+6\)

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

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

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

d) \(3x^3+2x-5\)

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

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

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

\(4x^3-7x^2+3x\)

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

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

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

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15\)

\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)-15\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-15\)

\(=\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)+1-16\)

\(=\left(x^2-5x+4+1\right)^2-4^2\)

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

\(=\left(x^2-4x+1\right)\left(x^2-4x+9\right)\)

a)\(x^3+4x^2-7x-10=x^3+x^2+3x^2+3x-10x-10=x^2\left(x+1\right)+3x\left(x+1\right)-10\left(x+1\right)\)

                                                   \(=\left(x+1\right)\left(x^2+3x-10\right)=\left(x+1\right)\left[\left(x^2+5x\right)-\left(2x+10\right)\right]=\left(x+1\right)\left(x+5\right)\left(x-2\right)\)

b) \(x^8+x+1=x^8-x^2+x^2+x+1=x^2\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

a) x³ - 7x - 6 = x³ + x² - x² - x - 6x - 6

= x²(x + 1) - x(x + 1) - 6(x + 1)

= (x + 1)(x² - x - 6)

= (x + 1)(x² + 2x - 3x - 6)

= (x + 1)[x(x + 2) - 3(x + 2)]

= (x + 1)(x + 2)(x - 3)