Bài 1: Tìm x, biết:
a) (x+3)^3 – x(3x+1)^2+(2x+1) (4x^2-2x+1)=28
b) x^2-1)^3-(x^4+x^2+1) (x^2-1)=0
Bài 1: Tìm x, biết: a) (x+3)^3 – x(3x+1)^2+(2x+1) (4x^2-2x+1)=28 b) x^2-1)^3-(x^4+x^2+1) (x^2-1)=0
By Reagan
By Reagan
Bài 1: Tìm x, biết:
a) (x+3)^3 – x(3x+1)^2+(2x+1) (4x^2-2x+1)=28
b) x^2-1)^3-(x^4+x^2+1) (x^2-1)=0
Đáp án:
a/ $x=0$ hoặc $x=-\dfrac{26}{3}$
b/ $x=0$ hoặc $x=1$ hoặc $x=-1$
Giải thích các bước giải:
a/ $(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28$
$⇔ x^3+9x^2+27x+27-x(9x^2+6x+1)+(8x^3+1)=28$
$⇔ x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-28=0$
$⇔ 3x^2+26x=0$
$⇔ x(3x+26)=0$
$⇔ \left[ \begin{array}{l}x=0\\x=-\dfrac{26}{3}\end{array} \right.$
b/ $(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0$
$⇔ (x^6-3x^4+3x^2-1)-(x^6-1)=0$
$⇔ x^6-3x^4+3x^2-1-x^6+1=0$
$⇔ -3x^2(x^2-1)=0$
$⇔ x^2(x-1)(x+1)=0$
$⇔ \left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.$
Đáp án:
a) `(x+3)^3 – x(3x+1)^2+(2x+1) (4x^2-2x+1)=28`
`<=>x^3+9x^2+27x+27-x(9x^2+6x+1)+((2x)^3+1^3)=28`
`<=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28`
`<=>(x^3+8x^3-9x^3)+(9x^2-6x^2)+27x-x=28-1=27`
`=>3x^2+26x=0`
`<=>x(3x+26)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x=3x+26=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-26}{3}\end{array} \right.\)
b) `(x^2-1)^3-(x^4+x^2+1) (x^2-1)=0`
`<=>x^6-3x^4+3x^2-1-(x^6-1)=0`
`<=>(x^6-x^6)+(1-1)+3x^2-3x^4=0`
`<=>3x^2.(1-x^2)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x^2=1\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `x={-1;0;1}`