288:
a: \(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-2x^3+x^2-x^3+2x^2-x+x^2-2x+1=0\)
=>\(\left(x^2-2x+1\right)\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
b: \(3x^4-13x^3+16x^2-13x+3=0\)
=>\(3x^4-3x^3+3x^2-10x^3+10x^2-10x+3x^2-3x+3=0\)
=>\(\left(x^2-x+1\right)\left(3x^2-10x+3\right)=0\)
=>\(3x^2-10x+3=0\)
=>\(3x^2-9x-x+3=0\)
=>3x(x-3)-(x-3)=0
=>(x-3)(3x-1)=0
=>\(\left[\begin{array}{l}x-3=0\\ 3x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac13\end{array}\right.\)
c: \(6x^4+5x^3-38x^2+5x+6=0\)
=>\(6x^4+6x^3-36x^2-x^3-x^2+6x-x^2-x+6=0\)
=>\(\left(x^2+x-6\right)\left(6x^2-x-1\right)=0\)
=>(x+3)(x-2)(3x+1)(2x-1)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\\ 3x+1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\\ x=-\frac13\\ x=\frac12\end{array}\right.\)
d: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
=>\(x^5+x^4+x^3+x^4+x^3+x^2+x^3+x^2+x+x^2+x+1=0\)
=>\(\left(x^2+x+1\right)\left(x^3+x^2+x+1\right)=0\)
=>\(\left(x^2+x+1\right)\left(x^2+1\right)\left(x+1\right)=0\)
=>x+1=0
=>x=-1
285:
a: \(\left(x+3\right)^4+\left(x+5\right)^4=16\) (1)
Đặt x+4=a
(1) sẽ trở thành: \(\left(a-1\right)^4+\left(a+1\right)^4=16\)
=>\(\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)
=>\(\left(a^2+1\right)^2-4a\left(a^2+1\right)+4a^2+\left(a^2+1\right)^2+4a\left(a^2+1\right)+4a^2=16\)
=>\(2\left(a^2+1\right)^2+8a^2=16\)
=>\(\left(a^2+1\right)^2+4a^2=8\)
=>\(a^4+6a^2+1-8=0\)
=>\(a^4+6a^2-7=0\)
=>\(\left(a^2+7\right)\left(a^2-1\right)=0\)
=>\(a^2-1=0\)
=>(a-1)(a+1)=0
=>(x+3)(x+5)=0
=>\(\left[\begin{array}{l}x+3=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=-5\end{array}\right.\)
b: Đặt x-2=a
Phương trình sẽ trở thành:
\(a^4+\left(a-1\right)^4=1\)
=>\(a^4+\left(a^2-2a+1\right)^2=1\)
=>\(a^4+\left(a^2+1\right)^2-4a\left(a^2+1\right)+4a^2=1\)
=>\(a^4-4a\left(a^2+1\right)+4a^2+a^4+2a^2=1-1=0\)
=>\(2a^4+6a^2-4a\left(a^2+1\right)=0\)
=>\(a^4+3a^2-2a\left(a^2+1\right)=0\)
=>\(a^4+3a^2-2a^3-2a=0\)
=>\(a\left(a^3-2a^2+3a-2\right)=0\)
=>\(a\left(a-1\right)\left(a^2-a+2\right)=0\)
=>a(a-1)=0
=>(x-2)(x-3)=0
=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)
c: \(\left(x+1\right)^4+\left(x-3\right)^4=82\)
=>\(\left(x^2+2x+1\right)^2+\left(x^2-6x+9\right)^2=82\)
=>\(\left(x^2+1\right)^2+4x\left(x^2+1\right)+4x^2+\left(x^2+9\right)^2-12x\left(x^2+9\right)+36x^2=82\)
=>\(\left(x^4+2x^2+1+x^4+18x^2+81\right)+4x\left(x^2+1\right)-12x\left(x^2+9\right)+40x^2=82\)
=>\(\left(2x^4+20x^2\right)+4x^3+4x-12x^3-108x+40x^2=0\)
=>\(2x^4-8x^3+60x^2-104x=0\)
=>\(2x\left(x^3-4x^2+30x-52\right)=0\)
=>\(x\left(x-2\right)\left(x^2-2x+26\right)=0\)
mà \(x^2-2x+26=\left(x-1\right)^2+25\ge25>0\forall x\)
nên x(x-2)=0
=>\(\left[\begin{array}{l}x=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
d: Đặt x-2,5=a
=>x-1,5=x-2,5+1=a+1
Phương trình sẽ trở thành: \(a^4+\left(a+1\right)^4=1\)
=>\(a^4+\left(a^2+2a+1\right)^2=1\)
=>\(a^4+\left(a^2+1\right)^2+4a\left(a^2+1\right)+4a^2=1\)
=>\(a^4+a^4+2a^2+1+4a\left(a^2+1+a\right)=1\)
=>\(2a^4+2a^2+4a\left(a^2+a+1\right)=0\)
=>\(a^4+a^2+2a\left(a^2+a+1\right)=0\)
=>\(a^4+a^2+2a^3+2a^2+2a=0\)
=>\(a^4+2a^3+3a^2+2a=0\)
=>\(a\left(a^3+2a^2+3a+2\right)=0\)
=>\(a\left(a+1\right)\left(a^2+a+2\right)=0\)
mà \(a^2+a+2=\left(a+\frac12\right)^2+\frac74\ge\frac74>0\forall a\)
nên a(a+1)=0
=>(x-2,5)(x-1,5)=0
=>\(\left[\begin{array}{l}x-2,5=0\\ x-1,5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2,5\\ x=1,5\end{array}\right.\)
