\(a^3+b^3+c^3-3abc\)

\(=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)

\(=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)\)

\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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

a3+b3+c3−3abca^3+b^3+c^3-3abca3+b3+c3−3abc

=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc

=(a+b)3+c3−(3a2b+3ab2+3abc)=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)=(a+b)3+c3−(3a2b+3ab2+3abc)

=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)

=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)

=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)

=(a+b+c)(a2+b2+c2−ab−ac−ab)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)=(a+b+c)(a2+b2+c2−ab−ac−ab)

\(x^4-6x^3+12x^2-14x+3\)

\(=\left(x^4-2x^3+3x^2\right)-\left(4x^3-8x^2+12x\right)+\left(x^2-2x+3\right)\)

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

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

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

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

\(=\left(x^2-2x+3\right)\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\)

\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)

\(=\left(x-3\right)\left(x-1-3\right)\)

\(=\left(x-3\right)\left(x-4\right)\)

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

\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)

\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

ta có: (x+y).(x+2y).(x+3y).(x+4y) + y⁴ 

= (x² +5xy +  4y² ).(x² + 5xy + 6y² ) + y⁴ 

= (x² + 5xy + 5y² - y² ).(x² + 5xy + 5y² + y² ) + y⁴

= (x² + 5xy + 5y² )² - y⁴ + y⁴ = (x² + 5xy + 5y²)²

=> đpcm

bình phương 2 vế rồi giải phương trình bậc 4 là ra

2x^2 - 5x + 5 = \(\sqrt{5x-1}\)

<=> (2x^2 - 5x + 5)^2 = (\(\sqrt{5x-1}\))^2

<=> 4x^4 - 20x^3 + 45x^2 - 50x + 25 = 5x - 1

<=> 4x^4 - 20x^3 + 45x^2 - 50x + 25 - 5x + 1 = 0

<=> 4x^4 - 20x^3 + 45x^2 - 55x + 26 = 0

<=> (4x^3 - 16x^2 + 29x - 26)(x - 1) = 0

<=> (4x^2 - 8x + 13)(x - 2)(x - 1) = 0

mà 4x^2 - 8x + 16 # 0 nên:

(x - 2)(x - 1) =0

=> x = 2; x = 1

vẽ hình hộ cái ko hiểu đề

Viết đề hản hoi cho tôi cái