Ta có: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)\)

\(\left(\sqrt{3}\right)^2=P+\frac{2\left(z+y+x\right)}{xyz}\) 

Mà x+y+z=xyz

=> P+2=3=>P=1

Vậy P=1

 2x² -6x -x +3-2x²-10x=0

-17x+3=0

-17x=-3

x=\(\frac{17}{3}\)

A B C D d M N I

a, Xét tam giác ADC có : MN // DC hay MI // DC 

Theo định lí Ta - lét ta có : \(\frac{MA}{MD}=\frac{IA}{IC}\)

b, Xét tam giác ABC có : AB // MN hay AB // IN 

Theo định lí Ta - lét ta có : \(\frac{BN}{NC}=\frac{IA}{IC}\)

mà \(\frac{MA}{MD}=\frac{IA}{IC}\)( cmt )

Suy ra : \(\frac{MA}{MD}=\frac{NB}{NC}\)

Ta có: \(B=\frac{x^2+1}{x^2-x+1}\)

    \(\Leftrightarrow B=\frac{3x^2+3}{3.\left(x^2-x+1\right)}\)

    \(\Leftrightarrow B=\frac{x^2+2x+1+2.\left(x^2-x+1\right)}{3.\left(x^2-x+1\right)}\)

    \(\Leftrightarrow B=\frac{\left(x+1\right)^2}{3.\left(x^2-x+1\right)}+\frac{2}{3}\)

Vì \(\frac{\left(x+1\right)^2}{3\left(x^2-x+1\right)}\ge0\forall x\)( Điều này các bạn tự CM nhé )

\(\Rightarrow\)\(B\ge\frac{2}{3}\)\(\Rightarrow\)\(B_{min}=\frac{2}{3}\)

Dấu "=" Xảy ra khi: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

Vậy \(B_{min}=\frac{2}{3}\)\(\Leftrightarrow\)\(x=-1\)

\(E=\frac{x-1}{x}:\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]\)

\(=\frac{x-1}{x}:\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-\frac{3x\left(x+1\right)}{3x}\right)\right]\)

\(=\frac{x-1}{x}:\left[\frac{2}{3x}-\frac{2}{x+1}.\frac{\left(x+1\right)\left(1-3x\right)}{3x}\right]\)

\(=\frac{x-1}{x}:\left[\frac{2}{3x}-\frac{2\left(x+1\right)\left(1-3x\right)}{3x\left(x+1\right)}\right]=\frac{x-1}{x}:\left[\frac{2}{3x}-\frac{2\left(1-3x\right)}{3x}\right]\)

\(=\frac{x-1}{x}:2=\frac{x-1}{2x}\)hay \(E=\frac{x-1}{2x}\)

mình làm ví dụ một câu thôi nhé ! 

\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)=\frac{4x}{\left(x+1\right)^2}\)

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

\(\frac{4x}{\left(x-1\right)\left(x+1\right)}:\frac{x-1+x^2+x+2}{\left(x+1\right)\left(x-1\right)}=\frac{4x}{\left(x+1\right)^2}\)

\(\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}=\frac{4x}{\left(x+1\right)^2}\)

<=> 4x( x + 1 )^2 = 4x ( x + 1 )^2 

<=> 4x ( x^2 + 2x + 1 ) = 4x ( x^2 + 2x + 1 )

<=> 4x^3 + 8x^2 + 4x - 4x^3 - 8x^2 - 4x = 0 

<=> 0 = 0 ( Vậy ta có đpcm ) 

Cảm ơn câu trả lời của bạn,mình quên không viết cap.Nhờ mọi người giải giúp mình bài 2 nhé!Bài 1 mình giải được rồi.Cảm ơn bạn lầu trên

Ta có: \(\left(x^2+1\right)^2+3x.\left(x^2+1\right)+2x^2=0\)

    \(\Leftrightarrow\left(x^2+1\right)^2+x.\left(x^2+1\right)+2x.\left(x^2+1\right)+2x^2=0\)

    \(\Leftrightarrow\left(x^2+1\right)\left(x^2+2x+1\right)+x.\left(x^2+2x+1\right)=0\)

    \(\Leftrightarrow\left(x^2+x+1\right)\left(x+1\right)^2=0\)

Vì  \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)mà \(\left(x^2+x+1\right)\left(x+1\right)^2=0\)

     \(\Rightarrow\left(x+1\right)^2=0\)\(\Leftrightarrow\)\(x=-1\)

Vậy \(x=-1\)