ĐK;x≥−1 pt ⇔(9x+12)√6x+8−(19x+26)√x+1≤√(x2+2x−3)3+√x2+2x−3 (1) Đặt √6x+8=a;√x+1=b⇒9x+12=a2+3b2+1;19x+26=3a2+b2+1 (1) ⇔(a2+3b2+1)a−(3a2+b2+1)b≤√(x2+2x−3)3+√x2+2x−3 ⇔(a−b)3+(a−b)≤√(x2+2x−3)3+√x2+2x−3 ⇔(a−b−√x2+2x−3)(a−b+(a−b)√x2+2x−3+x2+2x−3+1) ⇔a−b≤√x2+2x−3 do (....)>0 ⇔√6x+8−√x+1≤√x2+2x−3 ⇔6x+8≤(√x2+2x−3+√x+1)2 ⇔(√x2−1−√x+3)(3√x+3+√x2−1)≥0 ⇔x2−1\geqx+3 ⇔x1−√172orx≥1+√172kh vs ĐK ⇒x≥1+√172
ĐK;
x≥−1 pt
⇔(9x+12)√6x+8−(19x+26)√x+1≤√(x2+2x−3)3+√x2+2x−3 (1) Đặt
√6x+8=a;√x+1=b⇒9x+12=a2+3b2+1;19x+26=3a2+b2+1 (1)
⇔(a2+3b2+1)a−(3a2+b2+1)b≤√(x2+2x−3)3+√x2+2x−3 ⇔(a−b)3+(a−b)≤√(x2+2x−3)3+√x2+2x−3 ⇔(a−b−√x2+2x−3)(a−b+(a−b)√x2+2x−3+x2+2x−3+1) ⇔a−b≤√x2+2x−3 do (....)>0
⇔√6x+8−√x+1≤√x2+2x−3 ⇔6x+8≤(√x2+2x−3+√x+1)2 ⇔(√x2−1−√x+3)(3√x+3+√x2−1)≥0 $ \Leftrightarrow x^{2}-1
\geq
x+3
\Leftrightarrow x
\leq \frac{1-\sqrt{17}}{2}or x\geq\frac{1+\sqrt{17}}{2}
khvsĐK\Rightarrow x\geq \frac{1+\sqrt{17}}{2}$