\(x^4+2x^3+7x^2+26x+37=\left(x^4+2x^3+2x^2+2x+x^2+1\right)+\left(4x^2+24x+36\right)\)
\(=\left(x^2+x+1\right)^2+4\left(x+3\right)^2\)
Đặt: \(x^2+x+1=A;x+3=B\)
\(A\left(A^2+4.B^2\right)=5B^3\Leftrightarrow\left(A^3+5A.B^2\right)-\left(A.B^2+5B^3\right)=0\)
\(\Leftrightarrow\left(A-B^2\right)\left(A^2+5B^2\right)=0\). Em làm tiếp nhé!
Vẫn chưa hiểu phân tích của cô Chi)):
Ta có: \(x^4+2x^3+7x^2+26x+37=\left(x^4+2x^3+2x^2+x^2+2x+1\right)\)
\(+\left(4x^2+24x+36\right)=\left(x^2+x+1\right)^2+4\left(x+3\right)^2\)
Đặt \(x^2+x+1=u;x+3=v\)
Phương trình trở thành \(u\left(u^2+4v^2\right)=5v^3\)
\(\Leftrightarrow u^3+4uv^2=5v^3\)
\(\Leftrightarrow\left(u^3-v^3\right)+\left(4uv^2-4v^3\right)=0\)
\(\Leftrightarrow\left(u-v\right)\left(u^2+uv+v^2\right)+4v^2\left(u-v\right)=0\)
\(\Leftrightarrow\left(u-v\right)\left(u^2+uv+5v^2\right)=0\)
+) \(u-v=0\Rightarrow u=v\)
\(\Rightarrow x^2+x+1=x+3\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)
+) \(u^2+uv+5v^2=0\)(vô nghiệm)
Vậy \(x=\pm\sqrt{2}\)
Cô chưa hiểu ý em! Em muốn phân tích: \(u^2+uv+5v^2=0\) vô nghiệm ???
\(u^2+uv+5v^2=0\Leftrightarrow u^2+2u\frac{v}{2}+\frac{v^2}{4}-\frac{v^2}{4}+5v^2=0\Leftrightarrow\left(u+\frac{v}{2}\right)^2+\frac{19}{4}v^2=0\)
<=> \(\hept{\begin{cases}u+\frac{v}{2}=0\\v=0\end{cases}}\Leftrightarrow u=v=0\)
u = 0 <=> \(x^2+x+1=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)vô lí.
Vậy \(u^2+uv+5v^2=0\)vô nghiệm.
