\(A=\frac{x}{3}+\frac{16}{3x}+\frac{15}{3}\ge2\sqrt{\frac{16x}{9x}}+\frac{15}{3}=\frac{23}{3}\)

Dấu ''='' xảy ra : <=> \(x=4\)

Vậy GTNN A = 23/3 <=> x = 4 

@phuongeieu : đề không cho x dương nên không thể xài Cauchy được nhé

\(G=\frac{2}{x^2+4y^2-4xy+4x-4y+10}\)

\(=\frac{2}{\left(x-2y\right)^2+2\left(x-2y\right)+1+2x+9}\)

\(=\frac{2}{\left(x-2y+1\right)^2+2x+9}\)

\(\frac{x^2+4x+4}{x^2-4}=\frac{x^2+3x+2}{A}\)

\(\Rightarrow A=\frac{\left(x^2-4\right)\left(x^2+3x+2\right)}{x^2+4x+4}=\frac{\left(x-2\right)\left(x+2\right)\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2}=\left(x-2\right)\left(x+1\right)\)

\(\frac{x^2+4x+4}{x^2-4}=\frac{x^2+3x+2}{A}\)

\(\Leftrightarrow\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+x+2x+2}{A}\)

\(\Leftrightarrow\frac{x+2}{x-2}=\frac{x\left(x+1\right)+2\left(x+1\right)}{A}\Leftrightarrow\frac{x+2}{x-2}=\frac{\left(x+2\right)\left(x+1\right)}{A}\)

\(\Leftrightarrow A\left(x+2\right)=\left(x+2\right)\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow A=\left(x-2\right)\left(x+1\right)\)

Ta có: 

\(\left(x-y\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Dấu \(=\)khi \(x=y=\frac{1}{2}\)

\(x^2-7x+2\left(x-7\right)=0\)

\(x\left(x-7\right)+2\left(x-7\right)=0\)

\(\left(x+2\right)\left(x-7\right)=0\)

\(x+2=0\)hoặc \(x-7=0\)

\(x=-2\)hoặc \(x=7\)

Vậy x=-2 hoặc x=7

Bài làm

x² - 7x + 2( x - 7 ) = 0

<=> ( x² - 7x ) + 2( x - 7 ) =0

<=> x( x - 7 ) + 2( x - 7 ) = 0

<=> ( x - 7 )( x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2\end{cases}}\)

Vậy x = 7 hoặc x = -2