a) \(5a^2-5ax-7a+7x\)

\(=5a\left(a-x\right)-7\left(a-x\right)\)

\(=\left(5a-7\right)\left(a-x\right)\)

c) \(x^2-\left(a+b\right).x+ab\)

\(=x^2-ax-bx+ab\)

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

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

a,\(5a^2-5ax-7a+7x=5a\left(a+x\right)-7\left(a+x\right)=\left(a+x\right)\left(5a-7\right)\)

a: =(x+y)^3+z^3-3xy(x+y)-3xyz

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)

\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

=(x^2+x+5)(x^2+x-2)

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

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

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

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

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

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

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

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

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

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

<=> \(a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)

<=> \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Tổng 3 số không âm bằng 0 <=> a=b=c=1

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

<=> \(a^2-ab+b^2-bc+c^2-ac=0\)

<=> \(2a^2-2ab+2b^2-2bc+2c^2-2ac=0\)

<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Tổng 3 số không âm bằng 0 <=> a=b=c

#NguyễnHoàngTiến ơi cảm ơn bạn đã giúp mình nhưng cho mình hỏi left với right trong bài của bạn có nghĩa là gì vậy hả, mình không hiểu lắm.

bc(a+d) 9b –c) – ac( b +d) (a-c) + ab(c+d) ( a-b)

                   = bc(a+d) [ (b-a) + (a-c)] – ac(a-c)(b+d) +ab(c+d)(a-b)

                   = -bc(a+d )(a-b) +bc(a+d)(a-c) –ac(b+d)(a-c) + ab(c+d)(a-b)

                   = b(a-b)[ a(c+d) –c(a+d)] + c(a-c)[ b(a+d) –a(b+d)]

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

                   = d(a-b)(a-c)(b-c)

=d(a-b)(c-a)(c-b)