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( 2x - 1 )( x - 3 ) - 2x( x + 5 ) = 0

⇔ 2x² - 7x + 3 - 2x² - 10x = 0

⇔ -17x + 3 = 0

⇔ x = 3/17

\(\left(2x-1\right)\left(x-3\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow2x^2-6x-x+3-2x^2-10x=0\)'

\(\Leftrightarrow-17x+3=0\Leftrightarrow x=\frac{3}{17}\)

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