a) Áp dụng hằng đẳng thức (x + y)³ = x³ + y³ + 3xy(x + y) ta có:

(a + b + c)³ - a³ - b³ - c³ = [(a + b) + c]³ - a³ - b³ - c³

= (a + b)³ + c³ + 3(a + b)c(a + b + c) - a³ - b³ - c³

= a³ + b³ + 3ab(a + b) + c³ + 3c(a + b)(a + b + c) - a³ - b³ - c³

= 3(a + b)(ab + ac + bc + c²)  = 3(a + b)[a(b + c) + c(b + c)]

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

A = 3x² - 7x + 8 = 3(x² - 7/3 + 49/36) + 47/12 = 3(x - 7/6)² + 47/12

Ta luôn có: 3(x - 7/6)² \(\ge\)0 \(\forall\)x

=> 3(x - 7/6)² + 47/12 \(\ge\)47/12 \(\forall\)x

Dấu "=" xảy ra khi: x - 7/6 = 0 <=> x = 7/6

Vậy Min của A = 47/12 tại x = 7/6

\(=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-28=0\)

\(=>3x^2+26x=0=>x\left(3x+26\right)=0\)

\(=>\orbr{\begin{cases}x=0\\3x+26=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}}\)

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

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

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

a) \(=>x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=0\)

\(24x+25=0=>x=-\frac{25}{24}\)

Mình nghĩ đề thiếu =0

Hok tốt !

là 1260 g nhé !!

~~ Học tốt!~~

\(A=\left(3x-4\right)^2-4\left(x+1\right)^2=0\)

\(\Rightarrow\left(3x-4\right)^2-2^2\left(x+1\right)^2=0\)

\(\Rightarrow\left(3x-4\right)^2-\left(2x-2\right)^2=0\)

\(\Rightarrow\left(3x-4-2x+2\right)\left(3x-4+2x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\5x-6=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{6}{5}\end{cases}}\)