a: \(9\left(x-y\right)^2-4x^2-8xy-4y^2=0\)
=>\(\left(3x-3y\right)^2-\left(2x+2y\right)^2=0\)
=>(3x-3y-2x-2y)(3x-3y+2x+2y)=0
=>(x-5y)(5x-y)=0
=>\(\left[{}\begin{matrix}x=5y\\y=5x\end{matrix}\right.\)
c: \(\left(x-2\right)^2+2\left(x^2-4\right)+\left(x+2\right)^2=0\)
=>\(\left(x-2\right)^2+2\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2=0\)
=>\(\left(x-2+x+2\right)^2=0\)
=>\(2x=0\)
=>x=0
e: \(x^2-3x-28=0\)
=>\(x^2-7x+4x-28=0\)
=>x(x-7)+4(x-7)=0
=>(x-7)(x+4)=0
=>\(\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
g: \(27x^3-9x^2-3x+1=0\)
=>\(9x^2\left(3x-1\right)-\left(3x-1\right)=0\)
=>\(\left(3x-1\right)\left(9x^2-1\right)=0\)
=>\(\left(3x-1\right)^2\cdot\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
`a) 9(x - y)^2 - 4x^2 - 8xy - 4y^2 = 0`
`<=> 9(x - y)^2 - 4(x^2 +2xy +y^2 ) = 0`
`<=> 9(x - y)^2 - 4(x+y)^2 = 0`
`<=> (3x - 3y)^2 - (2x + 2y)^2 = 0`
`<=> (3x - 3y + 2x + 2y) (3x - 3y - 2x - 2y) = 0`
`<=> (5x - y ) (x - 5y) = 0`
`<=> 5x = y` hoặc `x = 5y`
Vậy ....
`c) (x - 2)^2 + 2(x^2 - 4) + (x+2)^2 = 0`
`<=> (x - 2)^2 + 2(x-2)(x+2) + (x+2)^2 = 0`
`<=> (x - 2 + x + 2)^2 = 0`
`<=> (2x)^2 = 0`
`<=> 2x = 0`
`<=> x = 0`
Vậy ...
`e) x^2 - 3x - 28 = 0`
`<=> (x^2 - 7x) + (4x - 28) = 0`
`<=> x(x-7) + 4(x-7) = 0`
`<=> (x+5)(x-7) = 0`
`<=> x + 5 = 0` hoặc `x - 7 = 0`
`<=> x = -5` hoặc `x = 7`
Vậy ...
`27x^3 - 9x^2 - 3x + 1 = 0`
`<=> (27x^3 - 9x^2) - (3x - 1)= 0`
`<=> 9x^2 (3x - 1) - (3x - 1)= 0`
`<=> (9x^2 -1)(3x - 1)= 0`
`<=> (3x+1)(3x - 1)^2= 0`
`<=> 3x + 1 = 0` hoặc `3x - 1 = 0`
`<=> x = -1/3` hoặc `x = 1/3`
Vậy ...
