Đặt x=b-c; y=c-a; z=a-b

=>x+y+z=b-c+c-a+a-b=0

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

=>\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

=>ayz+bxz+cxy=0

=>\(a=\frac{(y-z)x^2}{y^2 + yz + z^2};\quad b=\frac{(z-x)y^2}{z^2 + zx + x^2};\quad c=\frac{(x-y)z^2}{x^2 + xy + y^2}\)

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

\(=\frac{1}{x^2}\cdot\frac{(y-z)x^2}{y^2 + yz + z^2}+\frac{1}{y^2}\cdot\frac{(z-x)y^2}{z^2 + zx + x^2}+\frac{1}{z^2}\cdot\frac{(x-y)z^2}{x^2 + xy + y^2}\)

\(=\frac{y-z}{y^2 + yz + z^2}+\frac{z-x}{z^2 + zx + x^2}+\frac{x-y}{x^2 + xy + y^2}\)

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

\(y^2 + yz + z^2 = (y + z)^2 - yz = (-x)^2 - yz = x^2 - yz\)

\(z^2 + zx + x^2 = y^2 - zx\)

\(x^2 + xy + y^2 = z^2 - xy\)

\(A=\frac{y-z}{x^2 - yz}+\frac{z-x}{y^2 - zx}+\frac{x-y}{z^2 - xy}\)

\(=\frac{(y-z)(y^2 - zx)(z^2 - xy) + (z-x)(z^2 - xy)(x^2 - yz) + (x-y)(x^2 - yz)(y^2 - zx)}{(x^2 - yz)(y^2 - zx)(z^2 - xy)}\)

=0

=>ĐPCM

Ta có: \(\frac{a+b}{c+d}=\frac{b+c}{a+d}\)

=>(a+b)(a+d)=(b+c)(c+d)

=>\(a^2+ad+ba+bd=bc+bd+c^2+cd\)

=>\(a^2+ad+ba=bc+c^2+cd\)

=>\(a^2-c^2+ad-cd+ba-bc=0\)

=>(a-c)(a+c)+d(a-c)+b(a-c)=0

=>(a-c)(a+c+b+d)=0

=>a-c=0

=>a=c

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

<=>\(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)

<=>c(a+b)(a+b+c)=-ab(a+b)

<=>(a+b)(ac+bc+c²)+ab(a+b)=0

<=>(a+b)(ac+bc+ab+c²)=0

<=>(a+b)(a+c)(c+b)=0

       a+b=0

<=> b+c=o

       c+a=0
 

Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0

Khi đó:

(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)

=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2

=a+b+2c+2|c|=a+b+2c+2|c|

Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c

Do đó:

(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b

⇒√a+c+√b+c=√a+b

Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0

Khi đó:

(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)

=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2

=a+b+2c+2|c|=a+b+2c+2|c|

Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c

Do đó:

(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b

⇒√a+c+√b+c=√a+b