ai bt làm ko chỉ mik với

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

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

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

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

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

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

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

<=>x^4 ₊ 2x³ + x² + 4x² +4x = 12

<=>   x⁴ + 2x³ + 5x²+ 10x - 6x - 12 =0

<=>     x³(x + 2 ) + 5x ( x+2) - 6 ( x +2 )=0

 <=> ( x + 2 ) ( x³ - x + 6x - 6 ) =0

<=>  ( x + 2 ) ( x ( x -1) ( x +1) + 6 ( x - 1)) = 0

<=>  ( x + 2 ) ( x - 1 ) ( x² + x+ 6 ) = 0

<=>   x + 2 = 0

<=>  x = -2

<=>   x - 1 = 0

<=> x = 1 

<=> x²  + x  = -6 ( vô nghiệm )