\(2x^3-10x^2+12x=2x\left(x^2-5x+6\right)=2x\left[\left(x^2-3x\right)-\left(2x-6\right)\right]=2x\left[x\left(x-3\right)-2\left(x-3\right)\right]=2x\left(x-2\right)\left(x-3\right)\)

\(\left(x-2\right)\left(x^2-2x+4\right)\)

\(x^4+2x^3+10x-25\)

\(=x^4+5x^2+2x^3+10x-5x^2-25\)

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

Ta có : 

đặt x bình ra ngoài

nhóm x^2 và 4/x^2 ; x và 2/x 

xong đặt ẩn phụ là ra nhé

Ta có : 

b) \(x^2y-x^3-10y+10x\)

\(=x^2\left(y-x\right)-10\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-10\right)\)

c) \(x^2\left(x-2\right)+49\left(2-x\right)\)

\(=\left(x-2\right)\left(x^2-49\right)\)

\(=\left(x-2\right)\left(x-7\right)\left(x+7\right)\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\\ =-3\left(x^2-2\cdot\dfrac{5}{3}x+\dfrac{25}{9}-\dfrac{10}{9}\right)\\ =\dfrac{10}{3}-3\left(x-\dfrac{5}{3}\right)^2\\ =\left[\sqrt{\dfrac{10}{3}}-\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\left[\sqrt{\dfrac{10}{3}}+\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\\ =\left(\dfrac{\sqrt{30}}{3}+\dfrac{5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}}{3}-\dfrac{5\sqrt{3}}{3}+x\sqrt{3}\right)\)

\(=\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\)

\(-3x^2+10x-5\)

\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{10}{9}\right)\)

\(=-3\left(x-\dfrac{5+\sqrt{10}}{3}\right)\left(x-\dfrac{5-\sqrt{10}}{3}\right)\)