\(\left(x^2+x\right)^2+4x^2+4x+3=0\)
<=>\(\left(x^2+x\right)^2+4\left(x^2+x\right)+3=0\)
đặt \(x^2+x=y\)ta được:
\(y^2+4y+3=0< =>y^2+y+3y+3=0< =>y\left(y+1\right)+3\left(y+1\right)=0< =>\left(y+1\right)\left(y+3\right)=0=>\left[{}\begin{matrix}y+1=0\\y+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}y=-1\left(1\right)\\y=-3\left(2\right)\end{matrix}\right.\)
giải (1) :
\(x^2+x=-1< =>x^2+x+1=0< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=0\)
=>\(\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\left(vônghieemj\right)\)
giải 2:
\(x^2+x=-3< =>x^2+x+3=0< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{11}{4}=0\)
<=>\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0< =>\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\left(vônghiemej\right)\)
vậy.................
chúc bạn hcoj tốt ^^
Cách khác:\(\left(x^2+x\right)^2+4x^2+4x+3=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+4x^2+4x+1+2=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(2x+1\right)^2+2=0\)(pt vô nghiệm)
Vậy ...
