(a+b)\(^2\)-(b-a)\(^2\)

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

\(=2a^2\)

a) (2x+3)²-2(2x+3)(2x+5)+(2x+5)²

=4x²+12x+9-(4x+6)(2x+5)+4x²+20x+25

=4x²+12x+9-(8x²+12x+20x+30)+4x²+20x+25

=4x²+12x+9-8x²-12x-20x-30+4x²+20x+25

=4

b) (x²+x+1)(x²-x+1)(x²-1)

=((x²+1)²-x²)(x²-1)

=(x⁴+x²+1)(x²-1)

=x⁶+x⁴+x²-x⁴-x²-1

=x⁶-1

c)(a+b-c)²+(a-b+c)²-2(b-c)²

=a²+b²+c²+2ab-2ac-2bc+a²+b²+c²-2ab+2ac-2bc-2(b²-2bc+c²)

=2a²+2b²+2c²-4bc-2b²+4bc-2c²

=2a²

d) (a+b+c)²+(a-b-c)²+(b-c-a)²+(c-a-b)²

= a²+b²+c²+2ab+2ac+2bc+a²+b²+c²-2ab-2ac+2bc+a²+b²+c²+2bc-2ab+2ac+a²+b²+c²-2ac-2bc+2ab

=4a²+4b²+4c²+4ab+4bc

d) (a+b+c)²+(a-b-c)²+(b-c-a)²+(c-a-b)²

= a²+b²+c²+2ab+2ac+2bc+a²+b²+c²-2ab-2ac+2bc+a²+b²+c²-2bc-2ab+2ac+a²+b²+c²-2ac-2bc+2ab

=4a²+4b²+4c²

Ta có: \(a^2b+b^2c+c^2a-ab^2-bc^2-a^2c\)

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

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

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

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

\(=\left(b-c\right)\left\lbrack a\left(a-b\right)-c\left(a-b\right)\right\rbrack=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

Ta có: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)\)

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

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

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

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

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

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

\(=\left(a-b\right)\left(b-c\right)\left\lbrack a^2b+a^2c-ac^2-bc^2\right\rbrack=\left(a-b\right)\left(b-c\right)\cdot\left\lbrack b\left(a^2-c^2\right)+ac\left(a-c\right)\right\rbrack\)

=(a-b)(b-c)(a-c)\(\left\lbrack b\left(a+c\right)+ac\right\rbrack\)

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

Ta có: \(C=\frac{a^2b+b^2c+c^2a-ab^2-bc^2-a^2c}{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}\)

\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+bc+ac\right)}\)

\(=\frac{1}{ab+bc+ac}\)

Tách ra mỗi câu một lần.

Dài quá không ai làm đâu.

Nhìn nản lắm.

Câu 3: 

a: \(49^2=2401\)

b: \(51^2=2601\)

c: \(99\cdot100=9900\)

a)(a+b+c)²-(a+b)²-(a+c)²-(b+c)²

=a²+b²+c²+2ab+2bc+2ca-a²-2ab-b²-a²-2ac-c²-b²-2bc-c²

=-a²-b²-c²

=-(a²+b²+c²)

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

=a²+2ab+b²+2bc+c²+2ac-a²-b²-c²+2ab+2bc-2ac+a²+b²+c²+2ab-2bc-2ac+a²+b²+c²-2ab-2ac+2bc

=2a²+2b²+2c²+4ab-4bc-4ac