Đáp án `+` Giải thích các bước giải:
` a) x^2(x-5) + 5 -x = 0 `
`-> x^2(x-5) + (5-x)= 0`
`-> x^2 (x-5) - (x-5) = 0`
`-> (x-5)(x^2 - 1)=0 `
`-> (x-5)(x+1)(x-1)=0`
`->` \(\left[ \begin{array}{l}x-5=0\\x - 1 = 0 ; x + 1 = 0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=5\\x=1 ; x = -1\end{array} \right.\)
Vậy ....
` b) x^2 (x+8) + x^2 = -8x `
`-> x^3 + 8x^2 + x^2 + 8x = 0`
`-> x^3 + 9x^2 + 8x = 0`
`-> x(x^2 + 9x + 8 ) = 0`
`-> x(x^2 + x + 8x +8) = 0`
`-> x[x(x + 1) + 8(x+1)] = 0`
`-> x(x+1)(x+8)=0`
`->` \(\left[ \begin{array}{l}x=0 ; x + 1 = 0\\x+8=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=0 ; x = -1\\x=-8\end{array} \right.\)
Vậy ....
` c) x^2 - 10x + 16 = 0`
`-> x^2 - 8x - 2x + 16 = 0`
`-> x(x - 8) - 2(x-8)=0`
`-> (x-8)(x-2)=0`
`->` \(\left[ \begin{array}{l}x-8=0\\x-2=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=8\\x=2\end{array} \right.\)
Vậy ....
` d) (2x - 1)^2 - 25 = 0`
`-> (2x-1)^2 - 5^2 = 0`
`-> [(2x-1) -5][(2x-1)+5] = 0`
`-> (2x - 1 -5)(2x - 1 + 5)=0`
`-> (2x - 6)(2x + 4 )=0 `
`->` \(\left[ \begin{array}{l}2x -6 = 0\\2x + 4 = 0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}2x = 6\\2x = -4\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy ...
