a) \(x^2+3x-x\sqrt{x^2+2}=1+2\sqrt{x^2+2}.\)
\(x^2+3x-1=\sqrt{x^2+2}\left(2+x\right)\)
\(\left(x^2+3x-1\right)^2=\left(x^2+2\right)\left(2+x\right)^2\)
\(x^4+6x^3+7x^2-6x+1=x^4+4x^3+6x^2+8x+8\)
\(2x^3+x^2-14x-7=0\)
\(\left(2x^3+x^2\right)-\left(14x+7\right)=0\)
\(x^2\left(2x+1\right)-7\left(2x+1\right)=0\)
\(\left(2x+1\right)\left(x^2-7\right)=0\)
đến đây bạn có thể tự làm
b)
\(10\sqrt{x^3+1}=3\left(x^2+2\right)\)
\(10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2-x+1+x+1\right)\)
\(10\sqrt{\left(X+1\right)\left(x^2-x+1\right)}-3\left(x^2-x+1\right)-3\left(X+1\right)=0\)
\(9\sqrt{\left(x^2-x+1\right)\left(x+1\right)}-3\left(x^2-x+1\right)-3\left(x+1\right)+\sqrt{\left(x^2-x+1\right)\left(x+1\right)}\)
\(3\sqrt{\left(x^2-x+1\right)}\left(3\sqrt{x+1}-\sqrt{x^2+x+1}\right)-\sqrt{x+1}\left(3\sqrt{x+1}-\sqrt{x^2-x+1}\right)\)
\(\left(3\sqrt{x+1}-\sqrt{x^2+x+1}\right)\left(3\sqrt{x^2-x+1}-\sqrt{x+1}\right)=0\)
\(\hept{\begin{cases}3\sqrt{x+1}-\sqrt{x^2+x+1}=0\\3\sqrt{x^2-x+1}-\sqrt{x+1}=0\end{cases}}\)
TH 1 :
\(9\left(x+1\right)=x^2+x+1\)
\(9x+9=x^2+x+1\)
\(x^2-8x-8=0\)
đến đây bạn có thể tự làm
TH2
\(9\left(x^2-x+1\right)=\left(x+1\right)\)
\(9x^2-9x+9-x-1=0\)
\(9x^2-10x-8=0\)
đến đây bạn có thể tự làm
