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Ta có : \(^{4x^2}\)+ \(\frac{1}{x^2}\)+ \(7\)\(=\)\(8x\)+ \(\frac{4}{x}\)
\(\Rightarrow\) \(4x^4\)+ \(1\)+ \(7x^2\)\(=\)\(8x^3\)+ \(4x\)
\(\Rightarrow\) \(4x^4\)- \(8x^3\)+ \(7x^2\)-\(4x\)+\(1\)\(=\)\(0\)
\(\Rightarrow\) 4x3(x-1) - 4x2(x-1) + 3x(x-1) - (x-1) = 0
\(\Rightarrow\) (x-1) ( 4x3 -4x2 +3x -1) = 0
\(\Rightarrow\) (x-1) [2x2(2x-1) - x(2x-1) + (2x-1)] = 0
\(\Rightarrow\) (x-1) (2x-1) (2x2 - x +1) = 0
\(\Rightarrow\) (x-1) (2x-1) [ 2(x-\(\frac{1}{4}\))2 + \(\frac{7}{8}\)] = 0
Dễ thấy : 2(x-\(\frac{1}{4}\))2 + \(\frac{7}{8}\)> 0 \(\forall x\)
\(\Rightarrow\) \(\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x\in\){1;\(\frac{1}{2}\)}

\(\left(x+3\right)^2=9\left(2x-1\right)^2\)
\(\Leftrightarrow x^2+6x+9=9\left(4x^2-4x+1\right)\)
\(\Leftrightarrow x^2+6x+9=36x^2-36x+9\)
\(\Leftrightarrow x^2+6x+9-36x^2+36x-9=0\)
\(\Leftrightarrow-35x^2+42x=0\)
\(\Leftrightarrow-7x\left(5x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-7x=0\\5x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{5}\end{cases}}}\)
\(\left(x+3\right)^2=9\left(2x-1\right)^2\)
\(\Leftrightarrow x^2+6x+9=9\left(4x^2-4x+1\right)\)
\(\Leftrightarrow x^2+6x+9=36x^2-36x+9\)
\(\Leftrightarrow-35x^2+42x=0\)
\(\Leftrightarrow-7x\left(5x-6\right)=0\Leftrightarrow x=0;\frac{6}{5}\)

ta có
\(\left(x+y+z\right)^2=3\left(xy+yz+xz\right)\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2-\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Leftrightarrow x=y=z\)

x2 - 5x = 0
<=> x( x - 5 ) = 0
<=> x = 0 hoặc x - 5 = 0
<=> x = 0 hoặc x = 5
x2 - 7x + 10 = 0
<=> x2 - 2x - 5x + 10 = 0
<=> x( x - 2 ) - 5( x - 2 ) = 0
<=> ( x - 2 )( x - 5 ) = 0
<=> x - 2 = 0 hoặc x - 5 = 0
<=> x = 2 hoặc x = 5
\(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow x=0;5\)
\(x^2-7x+10=0\Leftrightarrow x^2-2x-5x+10=0\)
\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow x=5;2\)

5x( x - y ) + 12x - 12y
= 5x( x - y ) + 12( x - y )
= ( x - y )( 5x + 12 )
x2 + 2xy - 9 + y2
= ( x2 + 2xy + y2 ) - 9
= ( x + y )2 - 32
= ( x + y - 3 )( x + y + 3 )
\(5x\left(x-y\right)+12x-12y\)
\(=5x\left(x-y\right)+12\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+12\right)\)
\(x^2+2xy-9+y^2\)
\(=\left(x^2+2xy+y^2\right)-9\)
\(=\left(x+y\right)^2-3^2\)
\(=\left(x+y-3\right)\left(x+y+3\right)\)