$a,TA-CÓ:\sqrt{(x^2-x-20)(x+1)}=\sqrt{(x+4)(x-5)(x+1)}=\sqrt{(x^2-4x-5)(x+4)}$ Đặt $:\sqrt{x^2-4x-5}=a;\sqrt{x+4}=b$ $pt\Leftrightarrow 2(x^2-4x-5)+3(x+4)=5\sqrt{(x+4)(x^2-4x-5)}\Leftrightarrow 2a^2+3b^2=5ab$dễ rồi nha
$a,TA-CÓ:\sqrt{(x^2-x-20)(x+1)}=\sqrt{(x+4)(x-5)(x+1)}=\sqrt{(x^2-4x-5)(x+4)}$.Đặt $:\sqrt{x^2-4x-5}=a;\sqrt{x+4}b$$pt\Leftrightarrow 2(x^2-4x-5)+3(x+4)=5\sqrt{(x+4)(x^2-4x-5)}\Leftrightarrow 2a^2+3b^2=5ab$dễ rồi nha
$a,TA-CÓ:\sqrt{(x^2-x-20)(x+1)}=\sqrt{(x+4)(x-5)(x+1)}=\sqrt{(x^2-4x-5)(x+4)}$
Đặt $:\sqrt{x^2-4x-5}=a;\sqrt{x+4}
=b$
$pt\Leftrightarrow 2(x^2-4x-5)+3(x+4)=5\sqrt{(x+4)(x^2-4x-5)}\Leftrightarrow 2a^2+3b^2=5ab$dễ rồi nha