A = \(\frac{1}{x^2-4x+9}\)= \(\frac{1}{x^2-4x+4+5}\)= \(\frac{1}{\left(x-2\right)^2+5}\)

Nhận xét :

( x - 2 ) ² > 0 với mọi x

=>  ( x - 2 ) ² + 5  > 5 

=> \(\frac{1}{\left(x-2\right)^2+5}\)< \(\frac{1}{5}\)

=> A < \(\frac{1}{5}\)

Dấu " = " xảy ra khi : ( x - 2 )² = 0

                              => x - 2 = 0

                               => x = 2

Vậy A _max = \(\frac{1}{5}\) khi x = 2

\(\frac{1}{\left(x^2-4x+9\right)}=\frac{1}{\left(x^2-2\cdot x\cdot2+4\right)+5}\)

\(=\frac{1}{\left(x-2\right)^2+5}\) mà \(\left(x-2\right)^2\ge0\)\(\Rightarrow\)\(\frac{1}{\left(x-2\right)^2+5}\ge\frac{1}{5}\)

Vậy Max A =1/5 khi x-2=0<->x=2

\(\left(x-3\right)^2+7x=9^2\)

=> \(x^2-6x+9+7x=81\)

=> \(x^2+x+9=81\)

=> \(x.\left(x+1\right)=72\)

=> \(x.\left(x+1\right)=8.9\)

=> \(x=8\)

Vậy x = 8

\(\left(x-3\right)^2+7x=9^2\)

 \(x^2-6x+3^2+7x=9^2\)

 \(x^2+\left(-6x+7x\right)+3^2=9^2\)

 \(x^2+x+3^2=9^2\)

 \(\left(x+3\right)^2=9^2\)

\(\Rightarrow\)\(x+3=9\) hoặc  \(x+3=-9\)

TH1:  \(x+3=9\Rightarrow x=9-3=6\)

TH2:  \(x+3=-9\Rightarrow x=-9-3=-12\)

Vậy \(x=6\)hoặc  \(x=-12\).

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