Question

giải phương trình

^{\(x^4+10x^3+26x^2+10x+1=0\)}

Answer 1

x4+10x3+26x2+10x+1=0x4+10x3+26x2+10x+1=0

⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0⇔x4+6x3+x2+4x3+24x2+4x+x2+6x+1=0

⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0⇔x2(x2+6x+1)+4x(x2+6x+1)+(x2+6x+1)=0

⇔(x2+4x+1)(x2+6x+1)=0⇔(x2+4x+1)(x2+6x+1)=0

⇔(x2+4x+4−3)(x3+6x+9−8)=0⇔(x2+4x+4−3)(x3+6x+9−8)=0

⇔[(x+2)2−3][(x+3)2−8]=0⇔[(x+2)2−3][(x+3)2−8]=0

⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2−3=0(x+3)2−8=0⇒[(x+2)2=3(x+3)2=8⇒[(x+2)2=3(x+3)2=8⇒⎡⎣⎢⎢⎢x=−4±12−−√2x=−6±32−−√2

Answer 2

Thử phân tích VT thành: \(\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\) xem sao?

Answer 3

\(x^4+10x^3+26x^2+10x+1=0\)

\(\Leftrightarrow\left(x^4+6x^3+x^2\right)+\left(4x^3+24x^2+4x\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow x^2\left(x^2+6x+1\right)+4x\left(x^2+6x+1\right)+\left(x^2+6x+1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+1\right)\left(x^2+4x+1\right)=0\)

\(\Leftrightarrow\left(x^2+6x+9-8\right)\left(x^2+4x+4-3\right)=0\)

\(\Leftrightarrow\left[\left(x+3\right)^2-8\right]\left[\left(x+2\right)^2-3\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2-8=0\\\left(x+2\right)^2-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=8\\\left(x+2\right)^2=3\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=\pm\sqrt{8}\\x+2=\pm\sqrt{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{8}-3\\x=\pm\sqrt{3}-2\end{cases}}}\)

Answer 4

thử sức với câu này xem sao

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