\(P=\frac{n-7+9}{n-7}=1+\frac{9}{n-7}\)
\(\left(\text{Để P}\right)max\Rightarrow\left(\frac{9}{n-7}\right)max\Rightarrow\left(n-7\right)min\text{ và }n-7>0\left(\text{vì }9>0\right)\)
n-7 min và n-7>0 => n-7=1 => n=8. Vậy MaxP=10
\(\hept{\begin{cases}b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\\c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\end{cases}}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}\)
áp dụng t.c dtsbn:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)