Câu 1
[tex]3a^2+3b^2=10ab \rightarrow 4(a^2-2ab+b^2)=a^2+2ab+b^2 \rightarrow 4(a-b)^2=(a+b)^2\rightarrow (\frac{a+b}{a-b})^2=4\rightarrow \frac{a+b}{a-b}=2(a>b\rightarrow a-b> 0)[/tex]
Vậy
[tex]P=\frac{a^2}{b(a+b)}+\frac{b^2}{a(b-a)}-\frac{a^2+b^2}{ab}=\frac{a^3(b-a)+b^3(a+b) +(a^2+b^2)(b^2-a^2)}{ab(b^2-a^2)}= \frac{ab(a^2+b^2)}{ab(b^2-a^2)}=\frac{a^2+b^2}{b^2-a^2}=\frac{\frac{10}{3}ab}{(b-a)(a+b)}=\frac{\frac{10}{3}ab}{\frac{-1}{2}(a+b)^2}=\frac{\frac{10}{3}ab}{\frac{-1}{2}(a^2+b^2+2ab)}=\frac{\frac{10}{3}ab}{\frac{-1}{2}(\frac{10}{3}ab+2ab)}=\frac{\frac{10}{3}ab}{\frac{-8}{3}ab}=\frac{5}{4}[/tex]
Vừa làm vừa kiểm tra lại nhé có thể mình có sai sót