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\(\dfrac{3a-2b}{5}\)=\(\dfrac{2c-5a}{3}\)=\(\dfrac{5b-3c}{2}\)=\(\dfrac{15a-10b}{5}\)=\(\dfrac{6c-15a}{9}\)=\(\dfrac{10b-6c}{2}\)

Suy ra: \(\dfrac{15a-10b+6c-15a+10b-6c}{25+9+4}\)=\(\dfrac{0}{38}\)=0

Suy ra: 3a=2b\(\Leftrightarrow\)\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)(1)

2c=5a\(\Leftrightarrow\)\(\dfrac{c}{5}\)=\(\dfrac{a}{2}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)

Theo tính chất của dãy tỉ số bằng nhau ta được:

\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)=\(\dfrac{a+b+c}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5

Tự làm nốt nha.

Đúng thì tick cho mk nha

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-5c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3+}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)

Mà \(a+b+c=-50\)

\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)

\(\Leftrightarrow2c=-50\)

\(\Leftrightarrow c=-25\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-15\end{matrix}\right.\)

Vậy ...

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}=\dfrac{5\left(3a-2b\right)+3\left(2c-5a\right)+2\left(5b-3c\right)}{25+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3a-2b=0\\2c-5a=0\\5b-3c=0\end{matrix}\right.\)
⇔ 15a= 10b = 6c ⇔ \(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)

\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)

Ta có:

\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)

\(+\frac{20b^3-\left(a^3+b^3\right)}{ab+5b^2}\le\frac{20b^3-ab\left(a+b\right)}{ab+5b^2}=\frac{b\left(20b^2-a^2-ab\right)}{b\left(a+5b\right)}=\frac{\left(4b-a\right)\left(a+5b\right)}{a+5b}=4b-a\)

( áp dụng bđt : \(a^3+b^3\ge ab\left(a+b\right)\) ( biến đổi tương đương là c/m đc ) )

Dấu "=" \(\Leftrightarrow a=b\)

+ Tương tự : \(\frac{19c^3-b^3}{bc+5c^2}\le4c-b\) Dấu "=" <=> b = c

\(\frac{19a^3-c^3}{ac+5a^2}\le4a-c\) Dấu "=" \(\Leftrightarrow a=c\)

Cộng vế theo vế ta có đpcm. Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{3}\)