Đáp án + Giải thích các bước giải:
`a,(x+2)(2x-5)=x^2+4x+4`
`⇔2x^2-5x+4x-10-x^2-4x-4=0`
`⇔x^2-5x-14=0`
`⇔(x^2+2x)-(7x+14)=0`
`⇔x(x+2)-7(x+2)=0`
`⇔(x+2)(x-7)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x-7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=7\end{array} \right.\)
Vậy `x∈{-2;7}`
`b,x^2-16=0`
`⇔(x-4)(x+4)=0`
`⇔` \(\left[ \begin{array}{l}x-4=0\\x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy `x∈{4;-4}`
`c,x^4-16=0`
`⇔(x^2-4)(x^2+4)=0`
`⇔(x^2-4)=0` `[ Do (x^2+4) > 0 ]`
`⇔(x-2)(x+2)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `x∈{2;-2}`
`d,x^2+5x+6=0`
`⇔(x^2+2x)+(3x+6)=0`
`⇔x(x+2)+3(x+2)=0`
`⇔(x+2)(x+3)=0`
`⇔` \(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy `x∈{-2;-3}`
`e,x^2-8x+12=0`
`⇔(x^2-6x)-(2x-12)=0`
`⇔x(x-6)-2(x-6)=0`
`⇔(x-6)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x-6=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=6\\x=2\end{array} \right.\)
Vậy `x∈{6;2}`
`f,x^4=1`
`⇔` \(\left[ \begin{array}{l}x^4=1^4\\x^4=(-1)^4\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `x∈{1;-1}`
