x4−30x2+31x−30
=x4+x−30x2+30x−30
=x(x3+1)−30(x2−x+1)
=x(x+1)(x2−x+1)−30(x2−x+1)
=(x2+x)(x2−x+1)−30(x2−x+1)
=(x2−x+1)(x2+x−30)
tự làm tieeps nhé
\(x^4-30x^2+31x-30=0\)
\(\Rightarrow x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30=0\)
\(\Rightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
\(\Rightarrow\left(x-5\right)\left[x^3+6x^2-x^2-6x+x+6\right]=0\)
\(\Rightarrow\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
Mà \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)
Chúc bạn học tốt.
x4 - 30x2 + 31x - 30 = 0
=> x4 - 5x3 + 5x3 - 25x2 - 5x2 + 25x + 6x - 30 = 0
=> x3 ( x - 5 ) + 5x2 ( x - 5 ) - 5x ( x - 5 ) + 6 ( x - 5 ) = 0
=> ( x - 5 ) ( x3 + 5x2 - 5x + 6 ) = 0
=> ( x - 5 ) ( x3 + 6x2 - x2 - 6x + x + 6 ) = 0
=> ( x - 5 ) [ x2 ( x + 6 ) - x ( x + 6 ) + ( x + 6 ) ] = 0
=> ( x - 5 ) ( x + 6 ) ( x2 - x + 1 ) = 0
Ta thấy :
x2 > x với mọi x thuộc R
=> x2 - x > 0
=> x2 - x + 1 > 1
=> Biểu thức trên vô nghiệm
Mà ( x - 5 ) ( x + 6 ) ( x2 - x + 1 ) = 0
=> \(\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Vậy \(x\in\left\{5;-6\right\}\)
