Uả , \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)mới đúng chứ ta...?

b1    \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

        \(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c;\frac{c}{a}=1\Rightarrow c=a\)

        \(\Rightarrow a=b=c\)

b2   \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}=k\)

                           =>  \(k=\frac{1}{2}\)                                       

1, \(A=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{x^2-1}\)

\(=\frac{9}{x+1}-\frac{8}{1-x}-\frac{16}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{9\left(1-x\right)\left(x-1\right)}{\left(x+1\right)\left(1-x\right)\left(x-1\right)}-\frac{8\left(x+1\right)\left(x-1\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}-\frac{16\left(1-x\right)}{\left(1-x\right)\left(x+1\right)\left(x-1\right)}\)

\(=\frac{9\left(1-x\right)\left(x-1\right)-8\left(x+1\right)\left(x-1\right)-16\left(1-x\right)}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)

\(=\frac{18x-9-9x^2-8x^2+8-16+16x}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17x^2+34x-17}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}\)

\(=\frac{-17\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)\left(1-x\right)}=\frac{-17\left(x-1\right)}{\left(x+1\right)\left(1-x\right)}\)

2, \(B=\frac{x^2+10x+25}{x+5}-\frac{x^2-6x+9}{x-3}\)

\(=\frac{\left(x+5\right)^2}{x+5}-\frac{\left(x-3\right)^2}{x-3}=x+5-x+3=8\)

\(2(a^2 -bc)^2 + 2(b^2 -ac)^2 +2(c^2-ab)^2 + (a^2-b^2)^2 + (b^2-c^2)^2 + (c^2-a^2)^2 >=0 \)

\(<=> 4a^4 + 4b^4 +4c^4 - 4.(a^2.bc + b^2.ac + c^2.ab) >=0 \)

\(<=> a^4+b^4+c^4>=abc(a+b+c)\)

\(<=> a^4+b^4+c^4>=abc\)   (đpcm)

Hok tốt !!!!!!!!!!!!!!!!!