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\(\frac{a.b}{a+b}=\frac{b.c}{b+c}=\frac{c.a}{c+a}\)
\(\Rightarrow\frac{a+b}{a.b}=\frac{b+c}{b.c}=\frac{c+a}{c.a}\) (vì a;b;c khác 0)
\(=\frac{a}{a.b}+\frac{b}{a.b}=\frac{b}{b.c}+\frac{c}{b.c}=\frac{c}{c.a}+\frac{a}{c.a}\)
\(=\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
=> a = b = c
\(P=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\frac{a.a^2+a.a^2+a.a^2}{a^3+a^3+a^3}=\frac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)
\(\frac{b}{a+b}=\frac{c}{b+c}=\frac{a}{a+c}\Rightarrow\frac{a+b}{b}=\frac{b+c}{c}=\frac{a+c}{a}\)
\(\Leftrightarrow\frac{a}{b}+1=\frac{b}{c}+1=\frac{c}{a}+1\)mà\(a,b,c>0\Rightarrow a+b+c\ne0\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Ta có:\(a^x=bc;b^y=ca;c^z=ab\Rightarrow a^xb^yc^z=a^2b^2c^2\)
\(\Leftrightarrow x;y;z=2\Rightarrow xyz=2.2.2=8=2+2+2+2=x+y+z+2\)
Do \(a^x=bc;b^y=ca;c^z=ab\Rightarrow a^x.b^y.c^z=bc.ca.ab=a^2.b^2.c^2\)\(\Leftrightarrow\frac{a^2.b^2.c^2}{a^x.b^y.c^z}=1\Rightarrow\frac{a^2}{a^x}.\frac{b^2}{b^y}.\frac{c^2}{c^z}=1\)
Do a;b;c;x;y;z>0;a;b;c>1\(\Rightarrow\hept{\begin{cases}\frac{a^2}{a^x}=1\\\frac{b^2}{b^y}=1\\\frac{c^2}{c^z}=1\end{cases}}\Rightarrow\hept{\begin{cases}a^2=a^x\\b^2=b^y\\c^2=c^z\end{cases}}\Rightarrow x=y=z=2\)
\(\Rightarrow\hept{\begin{cases}x+y+z+2=2+2+2+2=4\\x.y.z=2.2.2=4\end{cases}}\Rightarrow x+y+z+2=xyz\)
Ta co a^2+c^2/b^2+a^2=c/b
=> (a^2+c^2) x b= (b^2+a^2) x c
=> a^2b+c^2b=b^2c+a^2c
=> bcb+c^2b=b^2c+bcc
=> b^2c+c^2b=b^2c+bc^2
Thay \(a^2=b.c\) Ta có
\(\frac{b.c+c^2}{b^2+b.c}=\frac{c.\left(b+c\right)}{b.\left(b+c\right)}=\frac{c}{b}\)(dpcm)
Thay a2 = b.c ta được:
\(\frac{c}{b}=\frac{b.c+c^2}{b.c+b^2}=\frac{c\left(b+c\right)}{b\left(b+c\right)}=\frac{c}{b}\)(đúng)
Vậy ta được đpcm
