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

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

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

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

= [ x ( x + 10 ) ] [ ( x+4 ) ( x+ 6) +128

=( x² + 10x ) ( x² +10x + 24 ) +128

dat : x² + 10x =a , ta co:

a ( a + 24 ) +128

=a²  + 24a +128

= (a + 12 )²  - 16

= ( a+ 12 -4 )  ( a + 12 + 4)

= ( a +8 ) ( a + 16 )

= ( x² + 10x +8 )( x² + 10x + 4)

b)x^4+5x^2-6

=x⁴-x³+x³-x²+6x²-6x+6x-6

=x³(x-1)+x²(x-1)+6x(x-1)+6(x-1)

=(x-1)(x³+x²+6x+6)

=(x-1)[x²(x+1)+6(x+1)]

=(x-1)(x+1)(x²+6)

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

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

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

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

Đặt \(t=x^2-11x+30\)

\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=t.\left(t-2x\right)-24x^2\)

\(=t^2-2xt-24x^2\)

\(=\left(t^2-2xt+x^2\right)-25x^2\)

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

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

\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)

Tham khảo nhé~

\(x^2-3xy-40y^2\)

\(=x^2+5xy-8xy-40y^2\)

\(=x\left(x+5y\right)-8y\left(x+5y\right)\)

\(=\left(x+5y\right)\left(x-8y\right)\)

a)x⁷+x⁵+1=x⁷+x⁶-x⁶+2x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+1

=x⁷-x⁶+x⁵-x³+x²+x⁶-x⁵+x⁴-x²+x+x⁵-x⁴+x³-x+1

=x²(x⁵-x⁴+x³-x+1)+x(x⁵-x⁴+x³-x+1)+1(x⁵-x⁴+x³-x+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

b)4x⁴-32x²+1=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

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

=(2x²-6x+1)(2x²+6x+1)

c)x⁶+27=(x²+3)(x²-3x+3)(x²+3x+3)

d)3(x⁴+x²+1)-(x²+x+1)

=3x⁴-3x³+2x²+3x³-3x²+2x+3x²-3x+2

=x²(3x²-3x+2)+x(3x²-3x+2)+1(3x²-3x+2)

=(x²+x+1)(3x²-3x+2)

e)bạn tự làm nhé

a) \(x^2-2xy+y^2-xz+yz\) 

= \(\left(x-y\right)^2-z\left(x-y\right)\)

= \(\left(x-y\right)\left(x-y-z\right)\)

b) \(x^3+9x^2-4x-36\)

= \(x^3-2x^2+11x^2-22x+18x-36\)

= \(x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\)

= \(\left(x-2\right)\left(x^2+11x+18\right)\)

= \(\left(x-2\right)\left(x^2+2x+9x+18\right)\)

= \(\left(x-2\right)\left(x+2\right)\left(x+9\right)\)

Chuc ban hoc tot

\(\Rightarrow3\left(x^3-8\right)-3x^3-3x=-30\\ \Rightarrow3x^3-24-3x^3-3x=-30\\ \Rightarrow-3x=-6\Rightarrow x=2\)

\(3\left(x-2\right)\left(x^2+2x+4\right)-3x\left(x^2+1\right)=-30\)

\(\Leftrightarrow3x^2-24-3x^3-3x=-30\)

\(\Leftrightarrow x=2\)

3(x - 2)(x² + 2x + 4) - 3x(x² + 1) = -30

<=> 3(x³ - 8) - 3(x³ - x) = -30

<=> 3(x³ - 8 - x³ + x) = -30

<=> x - 8 = \(\dfrac{-30}{3}\)

<=> x - 8 = 10

<=> x = 10 - 8 

<=> x = 2