\(x^4-4x^3-x^3+4x^2+4x^2-4x-x+1=0\)0
\(x^3\left(x-4\right)-x^2\left(x-4\right)+4x\left(x-1\right)-\left(x-1\right)\)=0
\(\left(x^3-x^2\right)\cdot\left(x-4\right)+\left(4x-1\right)\cdot\left(x-1\right)=0\)
\(x^2\left(x-1\right)\cdot\left(x-4\right)+\left(4x-1\right)\cdot\left(x-1\right)=0\)
\(\left(x-1\right)\cdot\left(x^3-4x^2+4x-1\right)=0\)
\(x=1\)
Phương trình đã cho có dạng:
\(ax^4+bx^3+cx^2+a=0\left(a\ne0\right)\)
Đặt \(x+\frac{1}{x}=y\) ta đưa phương trình về dạng:\(y^2-5y+6=0\)
Giải phương trình bậc hai theo y ta có:\(y_1=2;y_2=3\)
Do đó:
\(x+\frac{1}{x}=2\Rightarrow x^2-2x+1=0\Rightarrow x_o=1\)
\(x+\frac{1}{x}=3\Rightarrow x^2-3x+1=0\Rightarrow x_1=\frac{3-\sqrt{5}}{2};x_2=\frac{3+\sqrt{5}}{2}\)
Vậy phương trình đã cho có ba nghiệm là:
\(x_o=1;x_1=\frac{3-\sqrt{5}}{2};x_2=\frac{3+\sqrt{5}}{2}\)(xo là nghiệm kép).
\(x^4-5x^3+8x^2-5x+1=0\)
\(\left(x^4-x^3\right)-\left(4x^3-4x^2\right)+\left(4x^2-4x\right)-\left(x-1\right)=0\)
\(x^3\left(x-1\right)-4x^2\left(x-1\right)+4x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^3-4x^2+4x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)\right]=0\)
\(\left(x-1\right)^2\left(x^2+x+1-4x\right)=0\)
\(\left(x-1\right)^2\left(x^2-3x+1\right)=0\)
\(\left(x-1\right)^2\left(x^2-2x.1,5+1,5^2-1,25\right)=0\)
\(\left(x-1\right)^2\text{ }\left[\left(x-1,5\right)^2-\left(\sqrt{1,25}\right)^2\right]=0\)
\(\left(x-1\right)^2\left(x-1,5-\sqrt{1,25}\right)\left(x-1,5+\sqrt{1,25}\right)=0\)
Đến đây bạn tự làm nốt nhé~
x^4 - 5x^3 + 8x^2 - 5x + 1= 0
(x^4 - x^3) - (4x^3 - 4x^2) + (4x^2 - 4x) - (x - 1) = 0
x^3 (x-1)-4x^2 (x-1)+4x (x-1)-(x-1)=0
(x-1)×(x^3-4x^2+4x-1)=0
(x-1)×[(x^3-x^2)-(3x^2-3x)+(x-1)]=0
(x-1)[x^2 (x-1)-3x (x-1)+(x-1)]=0
(x-1)×(x-1)×(x^2-3x+1)=0
