cái này dễ mà

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

= (2x-1)^3

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

\(\)

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)

\(=9x^2+18x+9-9x^2+12x-4\)

\(=30x+5\)

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

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

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

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

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

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

=\(\left(9x^2+18x+9\right)-\left(9x^2-12x+4\right)\)

=\(9x^2+18x+9-9x^2+12x-4\)

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

MHƯ VẬY ĐÚNG KHÔNG

Bạn trình bầy rõ ràng chút đi. Mk chẳng hiểu gì cả.

\(-25x^6-y^8+10x^3y^4\\ =-\left(25x^6-10x^3y^4+y^8\right)\\ =-\left[\left(5x^3\right)^2-2\cdot5x^3\cdot y^4+\left(y^4\right)^2\right]\\ \\ =\left(5x^3-y^4\right)^2\)

(x² + 2.x.3 + 3² - 1).(x² + 2.x.4 + 16 - 1) - 24

=[(x+3)² - 1]. [(x+4)²-1] -24

=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24

=(x+4)(x+2)(x+5)(x-3) - 24

(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)

\(=x^4-16x^2+100=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)

\(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)