Gọi độ dài ba cạnh của tam giác đó lần lượt là \(a,b,c\).

Ba chiều cao tương ứng lần lượt là \(h_a,h_b,h_c\).

Ta có: 

\(\frac{h_a+h_b}{2}\div\frac{h_b+h_c}{2}\div\frac{h_c+h_a}{2}=5\div7\div8\)

\(\Leftrightarrow\frac{h_a+h_b}{5}=\frac{h_b+h_c}{7}=\frac{h_c+h_a}{8}=\frac{2\left(h_a+h_b+h_c\right)}{5+7+8}=\frac{h_a+h_b+h_c}{10}=t\)

suy ra \(h_a+h_b=5t,h_b+h_c=7t,h_c+h_a=8t,h_a+h_b+h_c=10t\)

\(\Rightarrow\hept{\begin{cases}h_a=3t\\h_b=2t\\h_c=5t\end{cases}}\)

Ta có: \(a.h_a=b.h_b=c.h_c\)

\(\Leftrightarrow a.3t=b.2t=c.5t\Leftrightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{6}\).

Giúp e với e đang cần gấp ạ

Theo bài ra ta có:
\(\dfrac{28.3+30.3+31x+32.5+35.4}{3+3+x+5+4}=31,5\\ \Rightarrow\dfrac{474+31x}{15+x}=31,5\\ \Rightarrow474+31x=31,5\left(15+x\right)\\ \Rightarrow474+31x=472,5+31,5x\\ \Rightarrow0,5x-1,5=0\\ \Rightarrow x=3\)

a, \(P=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)x^2y+\left(1+\dfrac{1}{2}\right)xy^2-6xy=\dfrac{3}{2}xy^2-6xy\)

bậc 3 

b, Thay x = 1/2 ; y = -1 ta được 

\(\dfrac{3}{2}.\dfrac{1}{2}.1-\dfrac{6.1}{2}\left(-1\right)=\dfrac{3}{4}+3=\dfrac{15}{4}\)

Áp dụng BĐT cô si :

\(\frac{a+\left(b+c\right)}{2}\ge\sqrt{a\left(b+c\right)}>0\)

\(\Rightarrow\frac{2}{a+b+c}\le\frac{1}{\sqrt{a\left(b+c\right)}}\Rightarrow\frac{a}{\sqrt{a\left(b+c\right)}}\ge\frac{2a}{a+b+c}\)

\(\Rightarrow\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)

\(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\)

tương tự : \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

\(=>\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge2\left(1\right)\)

Do 2 > 1 nên đpcm