Phan tich da thuc sau thanh nhan tu:
(a+b+c)³–a³–b³–c³
Ai jup dung mik tik cho
phan tich da thuc sau thanh nhan tu : ab(a-b)+bc(b-c)+ca(c-a)
ab(a-b)+bc(b-c)+ac(c-a)
=ab(a-b)-bc[(a-b)+(c-a)] +ac(c-a)
=ab(a-b) -bc(a-b) -bc(c-a) + ac(c-a)
=b(a-c)(a-b) -c(a-c)(a-b)
=(b-c)(a-c)(a-b)
phan tich da thuc sau thanh nhan tu (bang cach dat nhan tu chung)
(4x-y)(a+b)(4x-y)(c-1)
\((4x-y)(a+b)(4x-y)(c-1)\)
\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)
\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)
(4x-y)(a+b)(4x-y)(c-1)
= ( 4x - y ) ( 4x - y ) = ( 4x - y ) 1 + 1 = ( 4y - 2 ) 2
= (a + b ) ( 4x - y )2 ( c - 1 )
Bài giải :
(4x-y)(a+b)(4x-y)(c-1)
= ( 4x - y ) ( 4x - y ) = ( 4x - y ) 1 + 1
= ( 4y - 2 ) 2
= (a + b ) ( 4x - y )2 ( c - 1 )
ab(a+b)+bc(b+c) phan tich da thuc thanh nhan tu
ab(a+b)+bc(b+c)=a2b+ab2+b2c+bc2=(.................không phân tích đc nhé
de dung day giup minh di ab(a+b)+bc(b+c)+ca(c+a)+3abc please
ab(a+b)+bc(b+c) phan tich da thuc thanh nhan tu
a^3(b-c)+ b^3(c-a)+c^3(a-b) phan tich da thuc thanh nhan tu
a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) phan tich da THUC THANH NHAN TU
phan tich da thuc thanh nhan tu a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(a^3-c^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(a^3-b^3\right)+\left(b^3-c^3\right)\right]+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
a(b3−c3)+b(c3−a3)+c(a3−b3)
=a(b3−c3)−b(a3−c3)+c(a3−b3)
=a(b3−c3)−b[(a3−b3)+(b3−c3)]+c(a3−b3)
=a(b3−c3)−b(b3−c3)−b(a3−b3)+c(a3−b3)
=(b3−c3)(a−b)−(a3−b3)(b−c)
=(b−c)(b2+bc+c2)(a−b)−(a−b)(a2+ab+b2)(b−c)
=(a−b)(b−c)[(b2+bc+c2)−(a2+ab+b2)]
=(a−b)(b−c)(bc+c2−a2−ab)
=(a−b)(b−c)[b(c−a)+(c−a)(c+a)]
=(a−b)(b−c)(c−a)(a+b+c)
phan tich da thuc thanh nhan tu a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)
phan tich da thuc thanh nhan tu abc+ab+ac+bc+a+b+c+1