Bài 1:
c) -Vì \(\left(a-3\right)^2+\left(3b+1\right)^{400}\ge0\)
\(\Rightarrow\left(a-3\right)^2+\left(3b+1\right)^{400}=0\)
\(\Rightarrow\left(a-3\right)^2=0;\left(3b+1\right)^{400}=0\)
\(\Rightarrow a=3;b=\dfrac{-1}{3}\)
\(P=28a^2b-9ab^2=28.3^2.\dfrac{1}{3}-9.3.\dfrac{1}{3^2}=81\)
Bài 2:
2) \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)
\(\left(b-a\right)\left(b+a\right)=b\left(b+a\right)-a\left(b+a\right)=b^2+ab-ab-a^2=b^2-a^2\)
\(\Rightarrow\dfrac{b^2-a^2}{a+b}=b-a\)
\(\Rightarrow\dfrac{b^2-a^2}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
\(\Rightarrow\dfrac{b^2-a^2}{a^2+ab}=\dfrac{b-a}{a}\)
\(\Rightarrow\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
