giải phương trình : 6x4 +7x3 -36x2 -7x +6 =0
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Chọn B
Ta có: B(x) = 6x4 - 7x3 + 6x2- 7x3 + 4x4 + 3 - 5x + 2x
= 10x4 - 14x3 + 6x2 - 3x + 3.
\(\orbr{\begin{cases}x^3=-1\\x^3=8\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-x^2+x^2-x-6x+6=0\)
\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+3x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)+3\left(x-2\right)\right]\left(x-1\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-1\right)=0\)
\(\Rightarrow x=\left\{-3;1;2\right\}\)
\(y^2-7y-8=0\Rightarrow\orbr{\begin{cases}y=-1\\y=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt[3]{-1}=-1\\x=\sqrt[3]{8}=2\end{cases}}\)
\(\dfrac{-6x^4+7x^3+5x+2}{3x+1}\)
\(=\dfrac{-6x^4-2x^3+9x^3+3x^2-3x^2-x+6x+2}{3x+1}\)
\(=\dfrac{-2x^3\left(3x+1\right)+3x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)}{3x+1}\)
\(=-2x^3+3x^2-x+2\)
6x4+7x3-36x2-7x+6=0
<=> 6x4-2x3+9x3-3x2-33x2+11x-18x+6=0
<=> 2x3(3x-1)+3x2(3x-1)-11x(3x-1)-6(3x-1)=0
<=> (3x-1)(2x3+3x2-11x-6)=0
<=>(3x-1)(2x3-4x2+7x2-14x+3x-6)=0
<=>(3x-1)[2x2(x-2)+7x(x-2)+3(x-2)]=0
<=>(3x-1)(x-2)(2x2+7x+3)=0
<=>(3x-1)(x-2)(2x2+6x+x+3)=0
<=>(3x-1)(x-2)[2x(x+3)+(x+3)]=0
<=>(3x-1)(x-2)(x+3)(2x+1)=0
th1: 3x+1=0 <=> x=\(-\frac{1}{3}\)
th2: x-2=0 <=> x=2
th3: x+3=0 <=> x=-3
th4: 2x+1=0 <=> x=-\(\frac{1}{2}\)