jup mình với mọi người

júp mink với ace ơi

a) \(-10x^2-17x+6\)

\(=-10x^2-20x+3x+6\)

\(=-10x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+2\right)\left(3-10x\right)\)

b)\(x^2-10x+9\)

\(=x^2-x-9x+9\)

\(=x\left(x-1\right)-9\left(x-1\right)\)

\(=\left(x-1\right)\left(x-9\right)\)

c) \(x^2-10x+24\)

\(=x^2-4x-6x+24\)

\(=x\left(x-4\right)-6\left(x-4\right)\)

\(=\left(x-4\right)\left(x-6\right)\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\\ =-3\left(x^2-2\cdot\dfrac{5}{3}x+\dfrac{25}{9}-\dfrac{10}{9}\right)\\ =\dfrac{10}{3}-3\left(x-\dfrac{5}{3}\right)^2\\ =\left[\sqrt{\dfrac{10}{3}}-\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\left[\sqrt{\dfrac{10}{3}}+\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\\ =\left(\dfrac{\sqrt{30}}{3}+\dfrac{5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}}{3}-\dfrac{5\sqrt{3}}{3}+x\sqrt{3}\right)\)

\(=\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\)

\(-3x^2+10x-5\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{10}{9}\right)\)

\(=-3\left(x-\dfrac{5+\sqrt{10}}{3}\right)\left(x-\dfrac{5-\sqrt{10}}{3}\right)\)

Ta có:

\(3x^3+10x^2+14x+8\)

\(=3x^3+4x^2+6x^2+8x+6x+8\)

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

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

x^2 - 10x + 24

= x^2  - 4x - 6x + 24 

= x(x - 4) - 6(x - 4)

= (x - 6)(x - 4)

ko vt lại đề

x²-6x-4x+24

=(x²-6x)-(4x-24)

=x(x-6)-4(x-6)

=(x-6)(x-4)

\(x^2-10x+24\)

\(\Leftrightarrow x^2-4x-6x+24\)

\(\Leftrightarrow(x^2-4x)-\left(6x-24\right)\)

\(\Leftrightarrow x\left(x-4\right)-6\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)\)

a) \(x^2-3x=x\left(x-3\right)\)

b) \(10x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(5x-4y\right)\)

c) \(x^2-9=\left(x-3\right)\left(x+3\right)\)

a) x²-3x=x(x-3)

c) x²-9=x²-3²=(x+3)(x-3)

a: \(x^2-3x=\left(x-3\right)\cdot x\)

c: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)

g. x⁴ - 16

<=> x⁴ - 4²

<=> (x²)² - 4²

<=> (x² - 4)(x² + 4)

i. -x² + 10x - 25

<=> -(x² - 10x + 25)

<=> -(x² -10x + 5²)

<=> -(x - 5)²

sao ko ai like tui, buồn

a: =4x^2+8x-3x-6

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

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

b: =3(3x^2-2x-1)

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

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

c: =2x^2-4x+x-2

=2x(x-2)+(x-2)

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

d: =3x^2+3x-2x-2

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

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

e: =3x^2+9x+x+3

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

=(x+3)(3x+1)

a) \(4x^2+5x-6\)

\(=4x^2+8x-3x-6\)

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

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

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

b) \(9x^2-6x-3\)

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

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

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

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

c) \(2x^2-3x-2\)

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

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

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

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

d) \(3x^2+x-2\)

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

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

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

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

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

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

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