1) a³ + b³ + c³ - 3abc

=(a + b)(a² - ab + b²) + c³ - 3abc

=(a + b)(a² - ab + b²) + c(a² - ab + b²) - 2abc - ca² - cb²

=(a + b + c)(a² - ab + b²) - (abc + b²c + bc² + ac² + abc + c²a) + c³ + ac² + bc²

=(a + b = c)(a² - ab + b²) - (a + b + c)(bc + ca) + c²(a + b + c)

=(a + b + c)(a² + b² + c² - ab - bc - ca)

2) \(\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)=\left(3a+5\right)\left(a-3\right)+2\left(7b-10\right)\left(1\right)\)

\(\Leftrightarrow3a^2+15a+2ab+10b-a-5-2ab+4b=3a^2+14a+15+14b-10\)

\(\Leftrightarrow3a^2+14a+14b-5=3a^2+14a+14b-5\)( đúng)

\(\Rightarrow\left(1\right)\) đúng (đpcm)

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

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

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

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

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

a: a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b)-3bac

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

=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

b: Đề sai rồi bạn

c: 2(a+b+c)*(b/2+c/2-a/2)

=(a+b+c)(b+c-a)

=(b+c)^2-a^2

=c^2+2bc+c^2-a^2

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

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

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

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

b) Bạn chỉ cần nhân bung cả 2 vế ra là được á .

c) \(2\left(a+b+c\right)\left(\dfrac{b}{2}+\dfrac{c}{2}-\dfrac{a}{2}\right)\)

\(=2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)

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

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

\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)

Giải:

Vì $a\in Z^{+}$

$\Rightarrow 5^b=a^3+3a^2+5>a+3=5^c$

$\Rightarrow 5^b>5^c\Rightarrow b>c$

$\Rightarrow 5^b⋮5^c$

$\Rightarrow a^3+3a^2+5⋮a+3$

$\Rightarrow a^2\left(a+3\right)+5⋮a+3$

Mà $a^2\left(a+3\right)⋮a+3$

$\Rightarrow 5⋮a+3$

$\Rightarrow a+3\in Ư\left(5\right)$

$\Rightarrow a+3\in \left\{\pm 1;\pm 5\right\}\left(1\right)$

Do $a\in Z^{+}\Rightarrow a+3\ge 4\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)$

$\Rightarrow a+3=5$

$\Rightarrow a=5-3$

$\Rightarrow a=2$$(\ast )$

Thay $(\ast )$ vào biểu thức ta có:

$2^3+{3.2}^2+5=5^b\iff b=2$

$2+3=5^c\iff c=1$

Vậy: $\{\begin{array}{c}a=2\\ b=2\\ c=1\end{array}$

a=2

b=2

c=1

tự giải mệt rồi 

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