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

\(P=2\left(a-b\right)^3-6ab\left(a+b\right)-3\left(a+b\right)^2+6ab\)

Mà : \(a+b=1\)

Nên : \(P=2.1^3-6ab+6ab-3.1^2=2-3=-1\)

Chứng mình : 

\(a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)

\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a-b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)

Mà a+b+c=0

\(=>0.\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab.0=0\Rightarrow0+0=0\)

0+0=0 đúng suy ra \(a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\))

T I C K cho mình nha ủng hộ nha cảm ơn bạn 

đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)

\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)

Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)

Ủng hộ = 1 cái t i c k nha cảm ơn

=>7x-3x=8

=>4x=8

=>x=8:4

=>=2

t i c k mình nha cảm ơn

\(7\cdot x=3\cdot x+8\)

\(\left(7-3\right).x=8\)

\(4.x=8\)

\(x=8:4=2.\)

 ( 2x - 1 )² - ( 2x + 5 )( 2x - 5 ) = 20

\(=4x^2+1-4x-\left(4x^2-25\right)=20\)

\(=4x^2+1-4x-4x^2+25=20\)

\(=26-4x=20\)

\(4x=26-20=6\)

\(x=\frac{6}{4}=1,5\)

432 là mẫu chung nha bạn

ủng hộ nha