Với \(x\ne\pm3\)ta có : \(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+3\right)\left(x-4\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)

\(=\frac{x^2-x-12-\left(x^2-4x+3\right)+21}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)

\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)

\(=\left(\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3-1}{x+3}\right)\)

\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}\div\frac{x+2}{x+3}\)

\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\div\frac{x+2}{x+3}\)

\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{x+2}\)

\(=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)

Đặt M = x² - 4xy + 5y² + 10x - 22y + 28

= (x² - 4xy + 4y²) + (10x - 20y) + 25 + (y² - 2y + 1) + 2

= (x - 2y)² + 10(x - 2y) + 25 + (y - 1)² + 2

= (x - 2y + 5)² + (y - 1)² + 2 \(\ge2\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy Min M = 2 <=> x = -3 ; y = 1

\(x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+\left(y^2-2y+1\right)+25+1\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+\left(y-1\right)^2+25+1\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu "=" xảy ra khi \(\Leftrightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy \(Min_A=2\Leftrightarrow x=3;y=1\)

ĐK: \(x\ne-2\)

Với x=0  không phải nghiệm của pt trên

Với \(x\ne0\)pt trên trở thành:

\(\frac{1}{x+\frac{4}{x}+4}+\frac{5}{x+\frac{4}{x}}=-2\) (*)

Đặt \(y=x+\frac{4}{x}+2\) pt trở thành \(\frac{1}{y+2}+\frac{5}{y-2}=-2\)

ĐK: \(y\ne\pm2\)

pt trở thành: \(y^2+3y=0\Leftrightarrow y\left(y+3\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y=-3\end{cases}}\)

+ Với y=0 thì \(x+\frac{4}{x}+2=0\Rightarrow x^2+2x+4=0\Rightarrow\left(x+1\right)^2+3=0\)( vn)

+ Với y=-3 thì \(x+\frac{4}{x}+2=-3\Rightarrow x^2+5x+4=0\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)( tmđk)

Vậy nghiệm của pt là x=-1; x=-4

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

\(ĐKXĐ:x\ne-2\)

\(\left(1\right)\Leftrightarrow\left(\frac{x}{x^2+4x+4}+1\right)+\left(\frac{5x}{x^2+4}+1\right)=0\)

\(\Leftrightarrow\frac{x^2+5x+4}{x^2+4x+4}+\frac{x^2+5x+4}{x^2+4}=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(\frac{1}{x^2+4x+4}+\frac{1}{x^2+4}\right)=0\)

\(\Leftrightarrow x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\x+4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\x=-4\end{cases}\left(tmđkxd\right)}}\)

( 7x + 2y )² + ( 7x - 2y )² - 2( 49x² - 4y² )

= ( 7x + 2y )² + ( 7x - 2y )² - 2( 7x - 2y )( 7x + 2y )

= [ ( 7x + 2y ) - ( 7x - 2y ) ]²

= ( 7x + 2y - 7x + 2y )²

= ( 4y )² = 16y²

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

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

\(=\left(4y\right)^2=16y^2\)

(x - 4)(x + 1) - (x² - 8x + 16) = 0

<=> (x - 4)(x + 1) - (x - 4)² = 0

<=> (x - 4)(x + 1 - x + 4) = 0

<=> (x - 4).5 = 0

<=> x - 4 = 0

<=> x = 4

\(\left(x-4\right)\left(x+1\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+1\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+1-x+4\right)=0\)

\(\Leftrightarrow5\left(x-4\right)=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)