1, 

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

<=> (a - 2)(b + 3) = (a + 2)(b - 3)

<=> ab + 3a - 2b - 6 = ab - 3a + 2b - 6

<=> 3a - 2b = -3a + 2b

<=> 6a = 4b

<=> 3a = 2b 

<=> \(\frac{a}{2}=\frac{b}{3}\)(Đpcm)

2,

Có:

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(=\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)

\(=\frac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}=0\)

=> bz - cy = 0

=> bz = cy

=> \(\frac{b}{y}=\frac{c}{z}\)(1)

=> cx - az = 0

=> cx = az

=> \(\frac{c}{z}=\frac{a}{x}\)(2)

Từ (1) và (2)

=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)(Đpcm)

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

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

\(=a-b+b-c+c-a=0\)

Từ đây ta suy ra được

\(\hept{\begin{cases}\frac{c^2}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}\le\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\\\frac{c^2}{a+b}+\frac{a^2}{b+c}+\frac{b^2}{c+a}\ge\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\end{cases}}\)

Dấu = xảy ra khi \(|a|=|b|=|c|\)

Cảm ơn bạn đã trả lời câu hỏi giúp mình

Ta có : \(\frac{a}{b}=\frac{b}{c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{b}{c}\right)^2=\frac{ab}{bc}\)(Áp dụng tính chất a = b => a² = b² = ab)
\(\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{ab}{bc}\)
\(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)(Trừ khử b trên tử và dưới mẫu còn a/c)

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

mà \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)

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

\(VT=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{-1}{a-c}+\frac{1}{a-b}+\frac{-1}{b-a}+\frac{1}{b-c}+\frac{-1}{c-b}+\frac{1}{c-a}\)

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

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

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

đề sai bạn ơi

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Leftrightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\)

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

\(=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)

\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 1 )
Tương tự,ta có:

\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-ba+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 2 )
\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+cb-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ( 3 )
Cộng vế theo vế của ( 1 );( 2 );( 3 ) suy ra đpcm