⇔(x2+2)[√x2−x+1−(x+2)]=−5x−3⇔(−5x−3) (x2+2√x2−x+1+x+2−1) =0⇒−5x−3=0⇒x=−35hoặc đỏ=0 ⇒x2−x=√x2−x+1 DK:x≤0;x≥1 ⇔x2−x+1−1=√x2−x+1Đặt a=xanh≥34 ⇔a2−a−1=0⇒a=1+√52;a=... (loại)⇒√x2−x+1=1+√52⇒x=1,87;x=−0,87
⇔(x2+2)[√x2−x+1−(x+2)]=−5x−3⇔(−5x−3) (x2+2√x2−x+1+x+2−1) =0⇒−5x−3=0⇒x=−35hoặc đỏ=0
⇒x2−x=√x2−x+1 DK:x≤0;x≥1 $\Leftrightarrow x^2-x+1-1=
\sqrt{x^2-x+1}
Đặta=xanh
\geq \frac{3}{4}
\Leftrightarrow a^2-a-1=0\Rightarrow a=\frac{1+\sqrt{5}}{2};a=...
(loại)\Rightarrow \sqrt{x^2-x+1}=\frac{1+\sqrt{5}}{2}$$\Rightarrow x=1,87;x=-0,87$