Ai giúp em với ạ, cần gấp (x+3) ³ – x(3x+1) ²+(2x+1)(4x ² – 2x+1)=28 10/07/2021 Bởi Madeline Ai giúp em với ạ, cần gấp (x+3) ³ – x(3x+1) ²+(2x+1)(4x ² – 2x+1)=28
`(x+3)^3-x.(3x+1)^2+(2x+1).(4x^2-2x+1)=28` `<=>(x+3)^3-x.(3x+1)^2+(2x+1).((2x)^2-2x.1+1^2)=28` `<=>x^3+3.x^2. 3+3.x.3^2+27-x.(9x^2+6x+1)+(2x)^3+1=28` `<=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28` `<=>(x^3-9x^3+8x^3)+(9x^2-6x^2)+(27x-x)+28=28` `<=>3x^2+26x=0` `<=>x.(3x+26)=0` `<=>`\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{-26}{3}\end{array} \right.\) Vậy `x=0` hoặc `x=-26/3` là giá trị cần tìm. [ Áp dụng:] `(a+b)^2=a^2+2ab+b^2` `(a+b)^3=a^3+3.a^2. b+3.a.b^2+b^3` `(a+b).(a^2-ab+b^2)=a^3+b^3` Bình luận
Đáp án+giải thích: $(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28$ ⇔$x^3+9x^2+27x+27-x(9x^2+6x+1)+8x^3+1=28$ ⇔$9x^3+9x^2+27x+28-9x^3-6x^2-x-28=0$ ⇔$3x^2+26x=0$ ⇔$x(3x+26)=0$ ⇔\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\3x=-26\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=\frac{-26}{3} \end{array} \right.\) $Vậy$ $x=0$ $hoặc$ $x=\frac{-26}{3}$ Học tốt nhé!!! Bình luận
`(x+3)^3-x.(3x+1)^2+(2x+1).(4x^2-2x+1)=28`
`<=>(x+3)^3-x.(3x+1)^2+(2x+1).((2x)^2-2x.1+1^2)=28`
`<=>x^3+3.x^2. 3+3.x.3^2+27-x.(9x^2+6x+1)+(2x)^3+1=28`
`<=>x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28`
`<=>(x^3-9x^3+8x^3)+(9x^2-6x^2)+(27x-x)+28=28`
`<=>3x^2+26x=0`
`<=>x.(3x+26)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{-26}{3}\end{array} \right.\)
Vậy `x=0` hoặc `x=-26/3` là giá trị cần tìm.
[ Áp dụng:] `(a+b)^2=a^2+2ab+b^2`
`(a+b)^3=a^3+3.a^2. b+3.a.b^2+b^3`
`(a+b).(a^2-ab+b^2)=a^3+b^3`
Đáp án+giải thích:
$(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28$
⇔$x^3+9x^2+27x+27-x(9x^2+6x+1)+8x^3+1=28$
⇔$9x^3+9x^2+27x+28-9x^3-6x^2-x-28=0$
⇔$3x^2+26x=0$
⇔$x(3x+26)=0$
⇔\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\3x=-26\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\frac{-26}{3} \end{array} \right.\)
$Vậy$ $x=0$ $hoặc$ $x=\frac{-26}{3}$
Học tốt nhé!!!