A=\(x^2+\left(3y\right)^2+4-6xy+12y-4x+x^2+4x+4+1996\)

A=\(\left(x-3y-2\right)^2+\left(x+2\right)^2+1996\ge1996\)

Vay gtnn cua A la 1996.Dat duoc khi va chi khi \(\hept{\begin{cases}x+2=0\\x-3y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\3y=-2-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}}\)

Study well

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

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

    = \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

 = \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

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

Chúc bạn học tốt !!!

GAME THÌ KO OK

\(\frac{2}{3^8}:\frac{-2}{3^7}=\frac{2}{3^8}.\frac{3^7}{-2}=-\frac{1}{3}\)

Chuc ban hoc tot

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