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1. \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{x^2-1}\)
= \(-\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\)
= \(\frac{-x-1+x-1+2}{\left(x-1\right)\left(x+1\right)}=0\)
c) \(\left(\frac{x^2-16}{x^2+8x+16}+\frac{6}{x+4}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16}{\left(x+4\right)^2}+\frac{6\left(x+4\right)}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\left(\frac{x^2-16+6x+24}{\left(x+4\right)^2}\right)\cdot\frac{2x}{x+2}\)
= \(\frac{x^2+6x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x-2}\)
= \(\frac{x^2+4x+2x+8}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}\)
= \(\frac{\left(x+4\right)\left(x+2\right)}{\left(x+4\right)^2}\cdot\frac{2x}{x+2}=\frac{2x}{x+4}\)
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)nên x = -y hoặc y = -z hoặc z = -x
-Nếu x = -y thì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{-y}+\frac{1}{y}+\frac{1}{z}=\frac{1}{z}\)
\(\frac{1}{x+y+z}=\frac{1}{-y+y+z}=\frac{1}{z}\)
Khi đó: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
Tương tự với các trường hợp y = -z và z = -x thì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
Théo bđt Cauchuy Schwarz dạng Engel ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{3}\)
Đặt a = y + z; b = z+ x; c = x+ y (a;b;c > 0)
=> x+ y + z = (a+b+c)/2
=> x= (a+b+c)/2 - a = (b+c- a)/2
y = (a+b+c)/2 - b = (a+c-b)/2; z = (a+b - c)/ 2
Khi đó \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\frac{b}{a}+\frac{c}{a}-1+\frac{a}{b}+\frac{c}{b}-1+\frac{a}{c}+\frac{b}{c}-1\right)\)
=> \(P=\frac{b+c-a}{2a}+\frac{a+c-b}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}.\left(\left(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)-3\right)\right)\)
AD BĐT Cô - si có: \(\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2;\frac{c}{a}+\frac{a}{c}\ge2\)
=> \(P\ge\frac{1}{2}.\left(2+2+2-3\right)=\frac{3}{2}\)=> Min P = 3/2
Dấu "=" khi a = b = c<=> x = y = z
Áp dụng Cauchy - Schwarz và AM-GM :
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{xz+yz}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\)
\(\ge\frac{\left(x+y+z\right)^2}{\frac{2\left(x+y+z\right)^2}{3}}=\frac{3}{2}\)
Đẳng thức xảy ra tại x=y=z
Đk: x,y,z khác 0.
ta có: \(\left(y-z\right)^2\ge0\Rightarrow y^2+z^2\ge2yz\Leftrightarrow x^2+y^2+z^2\ge x^2+2yz\Leftrightarrow\frac{yz}{x^2+2yz}\ge\frac{yz}{x^2+y^2+z^2}\)
tương tự thì \(A\ge\frac{xy}{x^2+y^2+z^2}+\frac{yz}{x^2+y^2+z^2}+\frac{xz}{x^2+y^2+z^2}=\frac{xy+yz+xz}{x^2+y^2+z^2}\)
từ đề bài =>\(\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
=> A =0