\(1,\\ a,=10x^2y\\ b,=x^2+7x\\ 2,\\ =x\left(3y+11z\right)\)

ủa phần a mình phân tích rồi mà bạn hu hu

\(=3x^2+6x-14x+2x^2\\ =5x^2-8x=x\left(5x-8\right)\)

\(a.\) \(ax^2-a^2x-x+a\)

\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)

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

\(=\left(ax-1\right)\left(x-a\right)\)

\(b.\) \(18x^3-12x^2+2x\)

\(=2x\left(9x^2-6x+1\right)\)

\(=2x\left(3x-1\right)^2\)

\(c.\) \(x^3-5x^2-4x+20\)

\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)

\(=x^2\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x^2-4\right)\left(x-5\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)

\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)

\(=x^2+15x+7x+105+15\)

\(=x^2+22x+120\)

\(=\left(x+10\right)\left(x+12\right)\)

Câu 2: A

Câu 1: B

a) \(2x\left(x-7\right)-5y\left(x-7\right)=\left(x-7\right)\left(2x-5y\right)\)

b) \(5x^3y+10x^2y+5xy=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(4y^2-4y-x^2+1=\left(2y-1\right)^2-x^2=\left(2y-1-x\right)\left(2y-1+x\right)\)

d) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)

a: \(=\left(x-7\right)\left(2x-5y\right)\)

b: \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

a) \(=\left(2x-5y\right)\left(x-7\right)\)

b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)

c) \(=\left(\left(4y^2-4y+1\right)-x^2\right)=\left(2y-1\right)^2-x^2=\left(2y-1+x\right)\left(2y-1-x\right)\)

Biểu thức này không phân tích thành nhân tử.