a) A=(4-5x)²-(3+5x)²=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

B=(3x-1)(1+3x)-(3x+1)²=9x²-1-(3x+1)²=9x²-1-(9x²+6x+1)=9x²-1-9x²-6x-1=-6x-2=-2(3x+1)

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: \(99\cdot100=9900\)

TA  có \(a^3+b^3+c^3\ge3abc\Rightarrow-a^3-b^3-c^3\le-3abc\)

Cần chứng minh \(a^2b+b^2c+c^2a+ca^2+bc^2+ab^2-3abc\ge0\)

\(=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(a+c\right)-3abc\)

\(\ge abc+abc+abc-3abc=0\)