Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\frac{2a+b}{c}=\frac{2b+c}{a}=\frac{2c+a}{b}=\frac{3\left(a+b+c\right)}{a+b+c}=3\)

\(\Rightarrow\hept{\begin{cases}2a+b=3c\\2b+c=3a\\3c+a=3b\end{cases}}\)

\(\Rightarrow BT=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{b}=6+\frac{1}{3}=\frac{19}{3}\)

Vậy nếu a+b+c = 0 thì sao ?

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\)

\(=\frac{2a+2b+2c}{a+b+c}=2\)

+ Từ \(\frac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow3a-2b=c\) và \(3a-c=2b\)

+ Tương tự ta cũng có \(3b-2c=a\) và \(3b-a=2c\)

Và \(3c-2a=b\); \(3c-b=2a\)

Thay vào P

\(P=\frac{c.a.b}{2.b.2.c.2.a}=\frac{1}{8}\)

cam on

cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh\(2a-\frac{3b}{a}+2b=2c-\frac{3d}{c}+2d\)

đề đúng không vậy ta ??

đề đúng

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{2a+b}{c}=\frac{2b+c}{a}=\frac{2c+a}{b}=\frac{2a+b+2b+c+2c+a}{a+b+c}=\frac{3b+3c+3a}{a+b+c}=3\)

=>2a+b=3c; 2b+c=3a; 2c+a=3b

\(\left(\frac{2a+b}{c}\right)+\left(\frac{a}{2b+c}\right)+\left(\frac{3b}{2c+a}\right)\)

\(=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{3b}=3+\frac13+1=4+\frac13=\frac{13}{3}\)