Question

a) Phân tích đa thức thành nhân tử: \(x^3-5x^2+8x-4\)

b) Tìm giá trị nguyên x để \(A⋮B\),biết:

\(A=10x^2-7x-5\)và B=2x-3

c) Cho x+y=1 và x,y khác 0. CMR:

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

Answer 1

a) \(x^3-5x^2+8x-4\)

\(=x^3-2x^2-3x^2+6x+2x-4\)

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

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

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

\(=\left(x-2\right)\left[x\left(x-1\right)-2\left(x-1\right)\right]\)

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

Answer 2

b) \(A=10x^2-15x+8x-12+7\)

\(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\)

\(A=\left(2x-3\right)\left(5x+4\right)+7\)

Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\)

Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\)

\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{2;1;5;-2\right\}\)

Vậy.......

Answer 3

Ta có:\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}=\frac{\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)-\left(x-y\right)}{x^3y^3-x^3-y^3+1}=\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{x^3y^3-\left(x+y\right)^3+3xy\left(x+y\right)+1}=\frac{\left(x-y\right)\left[\left(x+y\right)^2-2xy-1\right]}{xy\left(x^2y^2+3\right)}\)\(=\frac{-2xy\left(x-y\right)}{xy\left(x^2y^2+3\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)

Do đó \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)(đpcm)