a) ĐK : y ≥ 0\(B=x^2-3x\sqrt{y}+2y=x^2-x\sqrt{y}-2x\sqrt{y}+2y=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)=\left(x-\sqrt{y}\right)\left(1-2\sqrt{y}\right)\)b) Với : \(x=\dfrac{1}{\sqrt{5}-2};y=\dfrac{1}{9+4\sqrt{5}}\)( TM) ; ta có :
\(\left(\dfrac{1}{\sqrt{5}-2}-\sqrt{\dfrac{1}{9+4\sqrt{5}}}\right)\left(1-2\sqrt{\dfrac{1}{9+4\sqrt{5}}}\right)=\left(\dfrac{1}{\sqrt{5}-2}-\sqrt{\dfrac{1}{\left(\sqrt{5}+2\right)^2}}\right)\left(1-2\sqrt{\dfrac{1}{\left(\sqrt{5}+2\right)^2}}\right)=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{\sqrt{5}+2}\right)\left(1-\dfrac{2}{\sqrt{5}+2}\right)\)Bạn tự quy đồng rồi tính ra nôt nhé.