\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow xy=-yz-zx;yz=-xy-zx;zx=-xy-yz\)

Ta có: x²+2yz=x²+yz+yz=x²+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)

Tương tự: \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)

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

\(=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{xy\left(x-y\right)-xz\left(x-y+y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\frac{xy\left(x-y\right)-xz\left(x-y\right)-xz\left(y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)\(=\frac{\left(xy-xz\right)\left(x-y\right)-\left(xz-yz\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\frac{x\left(y-z\right)\left(x-y\right)-z\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=1\)

Ta có:

\(xy+yz+zx=4xyz\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

\(P=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)

\(\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{2}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)

\(\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)

áp dụng cô si sháp cho 4 số ta được :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)  Luôn đúng , ( tự chứng minh )

\(\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\frac{1}{a+b+c+d}\) luôn luôn đúng

áp dụng vào  P ta được như sau

\(\frac{1}{x+x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) luôn đúng :))

\(\frac{1}{x+y+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\frac{1}{x+y+z+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)

Cộng tất cả vào ta được

\(P\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)\Leftrightarrow P\le\frac{1}{4}\left(x+y+z\right)\)

Thèo đề \(xy+yz+xz=4xyz\Leftrightarrow xy+yz+xz=xyz+xyz+xyz+xyz\)

Tao cũng éo hiểu tại sao nó = nhau được

1 đề sai  , 2 tao sai thế thôi

Từ đề <=>\(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+zy}\Leftrightarrow xz=xy=zy\)

Có : \(zx=xy\Rightarrow y=z\left(\text{Vì }x\ne0\right),xy=zy\Rightarrow x=z\)

=> x=y=z 

tự tính M :]]

bạn nào t-i-k sai cho tớ làm lại hộ ạ :)

ai t-i-k sai thì làm lại đi :(( 

Bạn tham khảo tại đây:

Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)

\(\Rightarrow yz+zx+xy=0\)

Ta có : \(x^2+2yz=x^2+yz+yz\)

                              \(=x^2+yz-zx-xy\)

                              \(=x\left(x-z\right)-y\left(x-z\right)\)

                              \(=\left(x-y\right)\left(x-z\right)\)

Tương tự : \(y^2+2xz=y^2+xz+xz\)

                                    \(=y^2+xz-xy-yz\)

                                    \(=y\left(y-x\right)+z\left(x-y\right)\)

                                    \(=\left(x-y\right)\left(z-y\right)\)

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

\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)  \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

Tính thế làm gì bạn ê