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

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

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~

a, x^4+6x^3+11x^2+6x+1 
= x^4 + 6x^3 + 9x² + 2x² + 6x + 1 
= x^4 + 9x² + 1 + 6x^3 + 2x² + 6x 
= x^4 + 9x² + 1² + 2.x².3x + 2.x².1 + 2.3x.1
= (x² + 3x + 1)²

Mình làm được ý a nên tk 1 tk

a, x²-7x-14y+2x

=x(x+2)-7(x-2y)

b, x³-4x²y+4xy²-25x

=x³-4x²y+4xy²-y³-25x+y³

=(x-y)³-25x+y³

a ) = x(x+2) - 7(x+2y)

b) =  -4 xy ( x-y) + (x^3-25x)  [ câu này mk , chaqcs là làm đúng đâu ]

x⁴ - 12x² + 12x - 9 = (x⁴ - 3x³) + (3x³ - 9x²) + (- 3x² + 9x) + (3x - 9)

= (x - 3)(x³ + 3x² - 3x + 3) = 0

a) x⁴-12x³+12x-9=(x⁴-3x³)+(3x³-9x²)-(3x²-9x)+(3x-9)=x³(x-3)+3x²(x-3)-3x(x-3)+3(x-3)

=(x-3)(x³+3x²-3x+3)

b)P(x)=o=>x-3=0 và x³=3x²-3x+3=0

                 =>x=3 và x=rỗng

=>x=3

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

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

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

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

a) x⁴ - 4x³ + 8x + 3                                                         b)x³+ 3x²-2

= x⁴ - (2x³ + 2x³ ) + (2x+6x) + 3 + 4x² - 4x²                               = x³ + 2x²+ x² - 2 + 2x - 2x

= x⁴ - 2x³ - x²  - 2x³+4x² + 2x - 3x² +6x + 3                        = ( x³ + 2x² -2x) +( x²+ 2x - 2)

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

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

c) x²-6x²+ 16

= (- 5)x² + 16

= - ( 5x² - 16)

a/ (x + 1)(x - 3)( x - √2 - 1)(x + √2 - 1)

b/ (x + 1)(x - √3 + 1)(x + √3 + 1)