1. x+3-|2x-1| $\leq$ 0
<=> |2x-1| $\geq$ x+3
<=> \(\left[ \begin{array}{l}2x-1\geq x+3\\2x-1\leq -x-3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x\geq 4\\3x\leq -2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x\geq 4\\x\leq\frac{-2}{3}\end{array} \right.\)
Vậy S= (-∞;$\frac{-2}{3}$] ∪ [4;+∞)
2. |2x-3|-3|x+1| $\geq$ 1
<=> $(2x-3)^{2}$-$[3(x+1)]^{2}$ $\geq$ 1
<=> $(2x-3)^{2}$-$(3x+3)^{2}$ $\geq$ 1
<=> (2x-3-3x-3)(2x-3+3x+3) $\geq$ 1
<=> (-x-6).5x $\geq$ 1
<=> $-5x^{2}$-30x-1 $\geq$ 0
<=> $\frac{-15-2√55}{5}$ $\leq$ $\frac{-15+2√55}{5}$
Vậy S= ($\frac{-15-2√55}{5}$;$\frac{-15+2√55}{5}$)
3. $(m+1)x^{2}$+2(m+1)x+m-1 $\leq$ 0
·$TH_{1}$: a=0 <=> m+1=0 <=> m=-1
*m=-1, bpt trở thành: -2 $\leq$ 0, ∀x∈R: đúng => nhận m=-1
·$TH_{2}$: a$\neq$0 <=> m+1$\neq$0 <=> m$\neq$-1
Để $(m+1)x^{2}$+2(m+1)x+m-1 $\leq$ 0, ∀x∈R
<=> $\left \{ {{a<0} \atop {Δ\leq 0}} \right.$
<=> $\left \{ {{m+1<0} \atop {[2(m+1)]^2-4(m+1)(m-1)\leq 0}} \right.$
<=> $\left \{ {{m<-1} \atop {4(m^2+2m+1)-4(m^2-1)\leq 0}} \right.$
<=> $\left \{ {{m<-1} \atop {4m^2+8m+4-4m^2+4\leq 0}} \right.$
<=> $\left \{ {{m<-1} \atop {8m+8\leq 0}} \right.$
<=> $\left \{ {{m<-1} \atop {m\leq-1}} \right.$
<=> m<-1
Hợp 2 TH => m∈(-∞;-1]
Nếu bạn thấy hay thì cảm ơn và đánh giá cho mk 5 sao nha!
