Đáp án+Giải thích các bước giải:
`a) x^3-8≥3x^2+6x+12`
`⇔ (x^3-8)-(3x^2+6x+12)≥0`
`⇔ (x-2)(x^2+2x+4)-3(x^2+2x+4)≥0`
`⇔ (x-2-3)(x^2+2x+4)≥0`
`⇔ (x-5)(x^2+2x+4)≥0`
mà `(x^2+2x+4)≥4>0∀x∈R`
`⇒ x-5≥0`
`⇔ x≥5`
Vậy bpt `a)` có nghiệm là: `x≥5`
`b) ((3x)/5-1/3)^2≤(x/5+2/3)^2`
`⇔((3x)/5-1/3)^2-(x/5+2/3)^2≤0`
`⇔ ((3x)/5-1/3)-x/5-2/3)((3x)/5-1/3)+x/5+2/3)≤0`
`⇔ ((2x)/5-1)((4x)/5+1/3)≤0`
`⇔` $\left \{ {{\dfrac{2x}{5}-1≤0} \atop {\dfrac{4x}{5}+\dfrac{1}{3}}≥0} \right.$ hoặc $\left \{ {{\dfrac{2x}{5}-1≥0} \atop {\dfrac{4x}{5}+\dfrac{1}{3}}≤0} \right.$
`+)`$\left \{ {{\dfrac{2x}{5}-1≤0} \atop {\dfrac{4x}{5}+\dfrac{1}{3}}≥0} \right.$ `⇔`$\left \{ {{2x-5≤0} \atop {12x+5≥0}} \right.$`⇔`$\left \{ {{x≤\dfrac{5}{2}} \atop {x≥\dfrac{-5}{12}}} \right.$`⇔`$\dfrac{-5}{12}$≤`x`≤$\dfrac{5}{2}$
`+)`$\left \{ {{\dfrac{2x}{5}-1≥0} \atop {\dfrac{4x}{5}+\dfrac{1}{3}}≤0} \right.$` ⇔`$\left \{ {{2x-5≥0} \atop {12x+5≤0}} \right.$`⇔`$\left \{ {{x≥\dfrac{5}{2}} \atop {x≤\dfrac{-5}{12}}} \right.$`⇔`$\dfrac{-5}{12}$≥`x`≥$\dfrac{5}{2}$(vô nghiệm)
Vậy bpt `b)` có nghiệm là: $\dfrac{-5}{12}$≤`x`≤$\dfrac{5}{2}$
`c)(x+2)/(x-2)-(x-2)/(x+2)>(8x-1)/(x^2-4)`
`⇔(x+2)/(x-2)-(x-2)/(x+2)-(8x-1)/(x^2-4)>0`
`⇔ ((x+2)^2-(x-2)^2-8x+1)/((x+2)(x-2))>0`
`⇔ (x^2+4x+4-x^2+4x-4-8x+1)/(x^2-4)>0`
`⇔ 1/(x^2-4)>0`
mà `1>0`
`⇒ x^2-4>0`
`⇔ x^2>4⇔-2>x>2`
Vậy bpt `c)` có nghiệm là: `-2>x`và `x>2`
`d)(2x-3)(5-2x)≤0`
`⇔`$\left \{ {{2x-3≥0} \atop {5-2x≤0}} \right.$ hoặc $\left \{ {{2x-3≤0} \atop {5-2x≥0}} \right.$
`+)`$\left \{ {{2x-3≥0} \atop {5-2x≤0}} \right.$`⇔`$\left \{ {{x≥\dfrac{3}{2}} \atop {x≥\dfrac{5}{2}}} \right.$`⇔x≥`$\dfrac{5}{2}$
`+)`$\left \{ {{2x-3≤0} \atop {5-2x≥0}} \right.$`⇔`$\left \{ {{x≤\dfrac{3}{2}} \atop {x≤\dfrac{5}{2}}} \right.$`⇔x≤`$\dfrac{5}{2}$
Vậy bpt `d)` có nghiệm là: `x≥`$\dfrac{5}{2}$ và `x≤`$\dfrac{5}{2}$
`e) (x-1/2)/(2-x)+1≥0`
`⇔(x-1/2+2-x)/(2-x)≥0`
`⇔ (3/2)/(2-x)≥0`
mà `3/2>0`
`⇔ 2-x > 0`
`⇔ -x > -2`
`⇔ x < 2`
Vậy bpt `e)` có nghiệm là: `x < 2`
`f) (2x-5)/(3-x)>3/2`
`⇔(2x-5)/(3-x)-3/2>0`
`⇔ (7/2x-(19)/2)/(3-x)>0`
`⇔ `$\left \{ {{\dfrac{7}{2}x-\dfrac{19}{2}>0} \atop {3-x>0}} \right.$`⇔` $\left \{ {{x>\dfrac{19}{7}} \atop {x<3}} \right.$ `⇔ (19)/7 < x <3`
`⇔ `$\left \{ {{\dfrac{7}{2}x-\dfrac{19}{2}<0} \atop {3-x<0}} \right.$`⇔` $\left \{ {{x<\dfrac{19}{7}} \atop {x>3}} \right.$ `⇔(19)/7 > x > 3` (vô nghiệm)
Vậy bpt `f)` có nghiệm là: ` (19)/7 < x <3`
