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

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

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

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

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

1) \(\left(a+b\right)^2\)

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

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

\(=a^2+2ab+b^2\left(dpcm\right)\)

2) \(\left(a-b\right)^3\)

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

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

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

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

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

`a)` 

`(a+b)^2`

`=(a+b)(a+b)`

`=a^2+ab+ab+b^2`

`=a^2+2ab+b^2`

`->` ĐPCM

`b)` `(a-b)^3`

`=(a-b)(a-b)(a-b)`

`=(a^2-2ab+b^2)(a-b)`

`=a^3-3a^2b+3ab^2-b^3`

`->` ĐPCM

a) a > b

⇒ 2a > 2b (nhân hai vế với 2 > 0)

⇒ 2a - 3 > 2b - 3 (cộng hai vế với -3)

b) a < b

⇒ -3a > -3b (nhân hai vế với -3 < 0)

⇒ -3a + 2 > -3b + 2 (1) (cộng hai vế với 2)

5 > 2

⇒ -3a + 5 > -3a + 2 (2) (cộng hai vế với -3a)

Từ (1) và (2) ⇒ -3a + 5 > -3b + 2

cau pha ca 2 ve ra di

ta có :

(a+b)³-(a-b)³= a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

=6a²b+2b³

=2b(3a²+b²)

vậy (a+b)³-(a-b)³=2b(3a²+b²)

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

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

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

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

\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)

\(\left\{{}\begin{matrix}A=5x^4-7x^2+4xy+y^2\\B=-9x^4-4xy-7y^2\end{matrix}\right.\)

\(A+B=5x^4-7x^2+4xy+y^2-9x^4-4xy-7y^2\)

\(A+B=\left(5x^4-9x^4\right)+\left(4xy-4xy\right)-\left(7y^2-y^2\right)-7x^2\)

\(A+B=-4x^4-6y^2-7x^2\)

Vì:

\(x^4\ge0\Rightarrow-4x^4\le0\)

\(\left\{{}\begin{matrix}6y^2\ge0\\7x^2\ge0\end{matrix}\right.\)

\(\Rightarrow-4x^4-6y^2-7x^2\le0\)

Vậy A và B không cùng dương

\(P=\dfrac{3a-b}{2a+15}+\dfrac{3b-a}{2b-15}\)

\(P=\dfrac{3a-b}{2a+a-b}+\dfrac{3b-a}{2b-a+b}\)

\(P=\dfrac{3a-b}{3a-b}+\dfrac{3b-a}{3b-a}\)

\(P=1+1=2\)