\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)+b^4[\left(c-b\right)-\left(a-b\right)]+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)+b^4\left(c-b\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)

\(=a^4\left(b-c\right)-b^4\left(b-c\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)

\(=\left(b-c\right)\left(a^4-b^4\right)-\left(a-b\right)\left(c^4-b^4\right)\)

\(=\left(b-c\right)\left(a^2-b^2\right)\left(a^2+b^2\right)-\left(a-b\right)\left(c^2-b^2\right)\left(c^2+b^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)+\left(a-b\right)\left(b-c\right)\left(c+b\right)\left(c^2+b^2\right)\)

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

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

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

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

\(=2x-1\)

b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

đặt ẩn phụ là ra

tích cho t đi

a) (2018x - 1) - 2019x (2018x - 1)=0

<=> (2018x - 1)(1 - 2019x)=0

<=> 2018x-1=0

1-2019x=0

<=> x=1/2018

x=1/2019

Câu 1

C2:

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

\(=x^2\left(2x^2+x-6+\frac{1}{x}+\frac{2}{x^2}\right)\)

\(=x^2\left[2\left(x^2+\frac{1}{x^2}\right)+\left(x+\frac{1}{x}\right)-6\right]\)

Đặt \(x+\frac{1}{x}=t\) \(\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

Thay :

\(=x^2\left[2\left(t^2-2\right)+t-6\right]\)

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

\(=2x^2\left(t^2+\frac{t}{2}-5\right)\)

\(=2x^2\left[\left(t+\frac{1}{2}\right)^2-\left(\frac{9}{4}\right)^2\right]\)

\(=2x^2\left(t+\frac{11}{4}\right)\left(t-\frac{7}{4}\right)\)

\(=2x^2\left(x+\frac{1}{x}+\frac{11}{4}\right)\left(x+\frac{1}{x}-\frac{7}{4}\right)\)

Câu 1: Tiếp đó : = (2x + 1)(x³ - 4x + x + 2)

= (2x + 1)[x(x² - 4) + (x + 2)]

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

= (2x + 1)[(x + 2)(x² - 2x + 1)]

= (2x + 1)(x + 2)(x - 1)²

Câu 3: (a² + b² - 5)² - 4(ab + 2)²

= (a² + b² - 5)² - (2ab+ 4)²

= (a² + b² - 5 - 2ab - 4)(a² + b² - 5 + 2ab + 4)

= [(a - b)² - 3²)][(a + b)² - 1]

= (a - b - 3)(a - b + 3)(a + b + 1)(a + b - 1)

\(a.=x^3+3x^2y+3x^2y+9xy^2+3xy^2+9y^3\)

    \(=x^2\left(x+3y\right)+3xy\left(x+3y\right)+3y^2\left(x+3y\right)\)

    \(=\left(x+3y\right)\left(x^2+3xy+3y^2\right).\)

\(b.=9x^3+3x^2y+9x^2y+3xy^2+3xy^2+y^3\)

    \(=3x^2\left(3x+y\right)+3xy\left(3x+y\right)+y^2\left(3x+y\right)\)

    \(=\left(3x^2+3xy+y^2\right)\left(3x+y\right)\).