Question

Giải các phương trình sau:

1) x⁴-8x³+11x²+8x-12=0

2) 4x³+x²+x-3=0

3)x⁵-5x⁴-2x³+17x²-13x+2=0

4) x⁴= 24x²-32

Answer 1

1) \(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2x-6x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x\left(x-2\right)-6\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\\x=6\end{matrix}\right.\)

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Answer 2

Answer 3

2) \(4x^3+x^2+x-3=0\)

\(\Leftrightarrow4x^3-3x^2+4x^2-3x+4x-3=0\)

\(\Leftrightarrow x^2\left(4x-3\right)+x\left(4x-3\right)+\left(4x-3\right)=0\)

\(\Leftrightarrow\left(4x-3\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow4x-3=0\left(\text{vì }x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\right)\)

\(\Leftrightarrow4x=3\Leftrightarrow x=\dfrac{3}{4}\)

Vậy ...