a)hình như đề sai
b)\(x^3+2x^2-7x+4=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(3x^2-3x\right)-\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+3x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+3x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-x+4x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[\left(x^2-x\right)+\left(4x-4\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)+4\left(x-1\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+4=0\\x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\x=1\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
Vậy x=-4 hay x=1
a) x4 - x3 + 2x2 - x + 1 = 0
=> (x4 + 2x2 + 1) - (x3 + x) = 0
=> (x2 + 1)2 - x(x2 + 1) = 0
=> (x2 + 1)(x2 - x + 1) = 0
=> \(\left[{}\begin{matrix}x^2+1=0\left(loại\right)\\x^2-x+1=0\end{matrix}\right.\)
=> (x2 - x + 1/4) + 3/4 = 0
=> (x - 1/2)2 + 3/4 = 0 (loại)
=> pt vô nghiệm
b) Ta có x3 + 2x2 - 7x + 4 = 0
=> (x3 - x) + (2x2 - 6x + 4) = 0
=> x(x2 - 1) + 2(x2 - 3x + 2) = 0
=> x(x - 1)(x + 1) + 2(x2 - 2x - x + 2) = 0
=> (x - 1)(x2 + x) + 2(x - 1)(x - 2) = 0
=> (x - 1)(x2 + x + 2x - 4) = 0
=> (x - 1)(x2 + 3x - 4) = 0
=> (x - 1)(x2 + 4x - x - 4) = 0
=> (x - 1)2(x + 4) = 0
=> \(\left[{}\begin{matrix}x-1=0\\x+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy ...
