\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz\) Thay x+y+z=0 vào

\(\Rightarrow0=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\) (1)

Ta có

\(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2\) (2)

Bình phương 2 vế của (1)

\(\left(x^2+y^2+z^2\right)^2=4\left(xy+yz+xz\right)^2\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2+2xy^2z+2xyz^2+2x^2yz\right)\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left[x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)\right]\)

Do x+y+z=0 nên

\(\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2\right)\)

\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{2}=2x^2y^2+2y^2z^2+2x^2z^2\) (3)

Thay (3) vào (2)

\(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+\dfrac{\left(x^2+y^2+z^2\right)^2}{2}\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)

phân tích đa thức thành nhân tử

refer

https://lazi.vn/edu/exercise/1000869/giai-phuong-trinh-x4-x2-6-0

ta cho x⁴ là x² ta có pt:

x²-x-6=0

\(\Rightarrow\left\{{}\begin{matrix}x_1=3\\x_2=-2\end{matrix}\right.\)

1)

a) \(=3x^2\left(x^2-1\right)-\left(x^3-1\right)+x^8-3x^4+3x^2-1\)

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

2) 

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-6\left(x^2+5x\right)+45\)

\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)-36+45\)

\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)+9=\left(x^2+5x-3\right)^2\)

Ta có: \(x^4-\left(a^2+1\right)x^2+4a^2=0\)(1)

Đặt \(b=x^2\)

Phương trình sẽ trở thành là: \(b^2-\left(a^2+1\right)b+4a^2=0\) (2)

\(\Delta=\left(a^2+1\right)^2-4\cdot1\cdot4a^2=a^4+2a^2+1-16a^2\)

\(=a^4-14a^2+1\)

Theo Vi-et, ta có: \(b_1+b_2=-\frac{b}{a}=a^2+1;b_1b_2=4a^2\)

Để (1) có nghiệm thì (2) phải có hai nghiệm không âm

=>\(b_1b_2\ge0;b_1+b_2\ge0;\Delta\ge0\)

=>\(\begin{cases}a^2+1\ge0\\ 4a^2\ge0\\ a^4-14a^2+1\ge0\end{cases}\Rightarrow a^4-14a^2+1\ge0\)

=>\(a^4-14a^2+49\ge48\)

=>\(\left(a^2-7\right)^2\ge48\)

=>\(\left[\begin{array}{l}a^2-7\ge4\sqrt3\\ a^2-7\le-4\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2\ge4\sqrt3+7\\ a^2\le7-4\sqrt3\end{array}\right.\)

=>\(\left[\begin{array}{l}a^2\ge\left(2+\sqrt3\right)^2\\ a^2\le\left(2-\sqrt3\right)^2\end{array}\right.\Rightarrow\left[\begin{array}{l}a\ge2+\sqrt3\\ a\le-2-\sqrt3\\ -2+\sqrt3\le a\le2-\sqrt3\end{array}\right.\)