Cho phép mình làm câu b trước :v
b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Rightarrow x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)
\(\Leftrightarrow-3x^2\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2\cdot\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1;x=-1\end{matrix}\right.\)
Vậy \(x_1=-1;x_2=0;x=1\)
Xử câu a nào :D
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Rightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow0+3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x+26=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=-26\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
Vậy \(x_1=0;x_2=-\dfrac{26}{3}\)
Ok con tê tê 
a) Ta có :
(x+3)3-x(3x+1)2+(2x+1)(4x2-2x+1)=28
<=> x3+9x2+27x+27-9x3-6x2-x+8x3-4x2+2x+4x2-2x+1=28
<=> 3x2+26x+28=28
<=> x(3x+26)=0
<=> x=0
x=\(\dfrac{-26}{3}\)
b) ta có :
(x2-1)3-(x4+x2+1)(x2-1)=0
<=> (x2-1) ((x2-1)2-(x4+x2+1))=0
<=> (x2-1)( x4-2x2+1)=0
<=> (x2-1)3=0
<=> x2-1=0
<=> x=1
x=-1
=))) tại dài quá nên mị chỉ viết ngắn thôi nếu đúng tick cho mị nhé :"> luv
