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

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

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

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

=(x+1)(x-2)(x+5)

a)

\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)

b) \(x^3-4x^2y+4xy^2-y^3\)

\(=x^3-3x^2y-x^2y+3xy^2+xy^2-y^3\)

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

\(=\left(x-y\right)^3-xy\left(x-y\right)\)

\(=\left(x-y\right)\left[\left(x-y\right)^2-xy\right]\)

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

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

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

Ta có:x(3x²+4x-7)=x[(3x²-3x)+(7x-7)]=x[3x(x-1)+7(x-1)]=x(x-1)(3x+7)

 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

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

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

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

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

b: \(5x^2+6xy+y^2\)

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

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

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

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

d: \(4x^2+12x+9-y^2\)

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

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