\(a,\) Rút gọn \(Q=\dfrac{x^3+y^3}{x^2-xy+y^2}.\dfrac{x+y}{x^2-y^2}\left(dkxd:x\ne y\right)\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2-xy+y^2}.\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y}{x-y}\)
Vậy \(Q=\dfrac{x+y}{x-y}\) với \(x\ne y\)
\(b,\) Thay \(x=\sqrt{7-4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
\(y=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
vào \(Q=\dfrac{x+y}{x-y}\) \(\Rightarrow P=\dfrac{\left(2-\sqrt{3}\right)+\left(\sqrt{3}-1\right)}{\left(2-\sqrt{3}\right)-\left(\sqrt{3}-1\right)}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1}{2-\sqrt{3}-\sqrt{3}+1}=\dfrac{2-1}{2+1-2\sqrt{3}}=\dfrac{1}{3-2\sqrt{3}}\)
Vậy \(P=\dfrac{1}{3-2\sqrt{3}}\)