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;\(\frac{1}{2}\)}

Ta có:

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\(\left(x+3\right)^2=9\left(2x-1\right)^2\)

\(\Leftrightarrow x^2+6x+9=9\left(4x^2-4x+1\right)\)

\(\Leftrightarrow x^2+6x+9=36x^2-36x+9\)

\(\Leftrightarrow x^2+6x+9-36x^2+36x-9=0\)

\(\Leftrightarrow-35x^2+42x=0\)

\(\Leftrightarrow-7x\left(5x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-7x=0\\5x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{5}\end{cases}}}\)

\(\left(x+3\right)^2=9\left(2x-1\right)^2\)

\(\Leftrightarrow x^2+6x+9=9\left(4x^2-4x+1\right)\)

\(\Leftrightarrow x^2+6x+9=36x^2-36x+9\)

\(\Leftrightarrow-35x^2+42x=0\)

\(\Leftrightarrow-7x\left(5x-6\right)=0\Leftrightarrow x=0;\frac{6}{5}\)

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 )

\(5x\left(x-y\right)+12x-12y\)

\(=5x\left(x-y\right)+12\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+12\right)\)

\(x^2+2xy-9+y^2\)

\(=\left(x^2+2xy+y^2\right)-9\)

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

\(=\left(x+y-3\right)\left(x+y+3\right)\)