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

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

\(=a^3+3a^2b+3ab^2+b^3\)

\(+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^3\)

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

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

1) Ta có: 2xy - x² - y² + 16

= -(x² - 2xy + y² - 16)

= -[(x - y)² - 16]

= -(x - y - 4)(x - y + 4)

2) x³ + 2x²y + xy² - 9x

= x(x² + 2xy + y² - 9)

= x[(x + y)² - 9]

= x(x + y  - 3)(x + y + 3)

3) x⁴ - 2x² = x²(x² - 2)

1. 2xy-x²-y²+16= -(x²-2xy+y²-16) = -(x²-2xy+y²)-16 = -(x-y)^2-16= (x+y)²-4²= (x+y-4).(x+y+4)

2. x³+2x²y+xy²-9x= (có sai đề không vậy?)

vế sau đâu đề cc gì vậy

a) \(3x^2-5x-8\)

\(=3x^2+3x-8x-8\)

\(=3x\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-8\right)\)

b) \(x^4+6x^3+9x^2-16\)

\(=\left(x^2+3x\right)^2-16\)

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

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

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

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

:> Easy 

@Ngọc Ly : \(A^2+2AB+B^2\)

\(\left(A+B\right)^2=A^2+2AB+B^2\)

2. Đổi : 12m/s = 43,2km/h

Thời gian đi 2/3 quãng đường đầu : t1 = s1/v1 = 2/3s / v1 = 2s/3v1

Thời gian đi 1/3 quãng đường còn lại : t2 = s2/v2 = 1/3s / v2 = s/3v2

Ta có : t1+t2=t

<=> 2s/3v1 + s/3v2 =t

<=> 2.90/3.40 + 90/3.43,2=t

<=> t =  x( tự tính)

Suy ra : vtb = s1+s2/t1+t2 = s/t = 90/x =ykm/h

đ/s: y :)) ( thông cảm , nhác tính )

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

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

=> a=b=c

b, \(0=\left(a+b+c\right)^3=a^3+b^3+c^3+6abc+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2\)

\(=a^3+b^3+c^3+6abc+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)

\(=a^3+b^3+c^3+6abc-3abc-3abc-3abc\)

\(\Rightarrow a^3+b^3+c^3=3abc\)