a) \(3x\left(x-4\right)+15=3x^2\)

\(\Leftrightarrow3x^2-12x+15-3x^2=0\)

\(\Leftrightarrow-12x+15=0\)

\(\Leftrightarrow x=\frac{5}{4}\)

b) \(x^2+y^2-2x+8y+17=0\)

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

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

Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)

a) \(x^2-5x+5y-y^2\)

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

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

b) \(2\left(x^2+1\right)^2-8x^2\)

\(=2\left[\left(x^2+1\right)^2-4x^2\right]\)

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

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

\(4x^2-49y^2+4x+1\)

\(=\left(4x^2+4x+1\right)-49y^2\)

\(=\left(2x+1\right)^2-\left(7y\right)^2\)

\(=\left(2x-7y+1\right)\left(2x+7y+1\right)\)

\(-x^3+9x^2-27x+27\)

\(=-\left(x^3-9x^2+27x-27\right)\)

\(=-\left(x-3\right)^3\)

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

\(=5\left(x^2-2xy^2+y^4\right)\)

\(=5\left(x-y^2\right)^2\)

a) ( x + 1/2 )² - ( x + 1/2 )( x + 6 ) = 8

⇔ ( x + 1/2 )[ ( x + 1/2 ) - ( x + 6 ) ] = 8

⇔ ( x + 1/2 )( x + 1/2 - x - 6 ) = 8

⇔ ( x + 1/2 ).(-11/2) = 8

⇔ x + 1/2 = -16/11

⇔ x = -43/22

b) ( x² + 2x )² - 2x² - 4x = 3

⇔ ( x² + 2x )² - 2( x² + 2x ) = 3

Đặt t = x² + 2x

bthuc ⇔ t² - 2t - 3 = 0

          ⇔ ( t² - 2t + 1 ) - 4 = 0

          ⇔ ( t - 1 )² - 2² = 0

          ⇔ ( t - 1 - 2 )( t - 1 + 2 ) = 0

          ⇔ ( t - 3 )( t + 1 ) = 0

          ⇔ ( x² + 2x - 3 )( x² + 2x + 1 ) = 0

          ⇔ ( x² - x + 3x - 3 )( x + 1 )² = 0

          ⇔ [ x( x - 1 ) + 3( x - 1 ) ]( x + 1 )² = 0

          ⇔ ( x - 1 )( x + 3 )( x + 1 )² = 0

          ⇔ x - 1 = 0 hoặc x + 3 = 0 hoặc x + 1 = 0

          ⇔ x = 1 hoặc x = -3 hoặc x = -1

thanks

b) \(ĐKXĐ:x\ne0\)

\(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\)

\(\Leftrightarrow x^3.\left(5x-2\right):2x^3=\frac{1}{2}\)

\(\Leftrightarrow\frac{5x-2}{2}=\frac{1}{2}\)\(\Leftrightarrow5x-2=1\)

\(\Leftrightarrow5x=3\)\(\Leftrightarrow x=\frac{3}{5}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{3}{5}\)

c) \(ĐKXĐ:x\ne2\)

\(\frac{x^4-2x^2-8}{x-2}=0\)\(\Rightarrow x^4-2x^2-8=0\)

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

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

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

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

Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+2\ge2\)

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

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

So sánh với ĐKXĐ ta thấy: \(x=-2\)thỏa mãn 

Vậy \(x=-2\)