\(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

\(=-a^4b^4\left[\left(4b\right)^2+2\cdot4\cdot3\cdot ab+\left(3a\right)^2\right]\)

\(=-a^4b^4\cdot\left(3a+4b\right)^2\)

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

i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)

\(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\\ =\left(x+3\right)\left(x^2-3x+9-4x\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

\(x^3-4x^2-12x+27\)

\(=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

\(x^3-4x^2-12x+27\) 

\(=x^3-3x^2-9x+27-3x-x^2\)

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

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

\(a^4+a^2+1\)

\(=a^4-a+a^2+a+1\)

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

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

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

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

a⁴ + a² +1 = a⁴ +2a² + 1 - a²

                  = (a⁴ +2a² + 1) - a²

                  = (a² + 1)² - a²

                      ^{= (}a² + 1 - a)(a² + 1 +a)

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

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

bn làm đ rùi, tui đ cho bn

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

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

Mình đã làm sai, cô Hoài Thương đúng rồi.

A=x¹⁴+x⁷+1

   =(x¹⁴+x¹³+x¹²)-(x¹³+x¹²+x¹¹)+(x¹¹+x¹⁰+x⁹)-(x¹⁰+x⁹+x⁸)+(x⁸+x⁷+x⁶)-(x⁶+x⁵+x⁴)+(x⁵+x⁴+x³)-(x³⁺x²+x)+(x²+x+1)

Đặt B=x²+x+1

=>A=x¹²B-x¹¹B+x⁹B-x⁸B+x⁶B-x⁴B+x³B-xB+B

=>A=B(x¹²-x¹¹+x⁹-x⁸+x⁶-x⁴+x³-x+1)

Thay B=x²+x+1 vào A là xong

dùng bơ du là ra chư mấy

Do lười đánh máy nên tớ mới đặt B khi làm khỏi đặt cũng dc cái này là toán lớp 8

a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)

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