Đáp án + Giải thích các bước giải:
`L=((a+b)/(\root{3}{a}+\root{3}{b})-\root{3}{ab}):(\root{3}{a}-\root{3}{b})^2`
`=((\root{3}{a^3}+\root{3}{b^3})/(\root{3}{a}+\root{3}{b})-\root{3}{ab}). 1/(\root{3}{a}-\root{3}{b})^2`
`=[((\root{3}{a}+\root{3}{b})(\root{3}{a^2}-\root{3}{ab}+\root{3}{b^2}))/(\root{3}{a}+\root{3}{b}]-\root{3}{ab}]. 1/(\root{3}{a}-\root{3}{b})^2`
`=(\root{3}{a^2}-\root{3}{ab}+\root{3}{b^2}-\root{3}{ab}). 1/(\root{3}{a}-\root{3}{b})^2`
`=(\root{3}{a^2}-2\root{3}{ab}+\root{3}{b^2}). 1/(\root{3}{a}-\root{3}{b})^2`
`=(\root{3}{a}-\root{3}{b})^2. 1/(\root{3}{a}-\root{3}{b})^2`
`=1`
`P=[\frac{4a-9a^-1}{2a^(1/2)-3a^(-1/2)}+\frac{a-4+3a^-1}{a^(1/2)-a^(-1/2)}]^2`
`=[\frac{a^-1(4a^2-9)}{a^(-1/2)(2a-3)}+\frac{1/a(a-3)(a-1)}{a^(-1/2)(a-1)}]^2`
`=[\frac{a^-1(2a-3)(2a+3)}{a^(-1/2)(2a-3)}+\frac{a-3}{a^(-1/2).a}]^2`
`=[\frac{a^-1(2a+3)}{a^(-1/2)}+\frac{a-3}{a^(1/2)}]^2`
`=[a^(-1/2)(2a+3)+a^(-1/2)(a-3)]^2`
`=[a^(-1/2)(2a+3+a-3)]^2`
`=(3a.a^(-1/2))^2`
`=(3a^(1/2))^2`
`=9a`
