2x+1/x +3≥ 3-5x/5 +4x+1/4
giải bpt
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NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 5 2020
1.
- Với \(x\ge\frac{1}{2}\Rightarrow2x-1\le x+2\Rightarrow x\le3\Rightarrow\frac{1}{2}\le x\le3\)
- Với \(x< \frac{1}{2}\Rightarrow1-2x\le x+2\Rightarrow3x\ge-1\Rightarrow x\ge-\frac{1}{3}\)
Vậy nghiệm của BPT là \(-\frac{1}{3}\le x\le3\)
2.
Để pt có 2 nghiệm trái dấu
\(\Leftrightarrow ac< 0\Leftrightarrow\left(m+2\right)\left(2m-3\right)< 0\Rightarrow-2< m< \frac{3}{2}\)
3.
\(5x-1>\frac{2x}{5}+3\Leftrightarrow5x-\frac{2x}{5}>4\Leftrightarrow\frac{23}{5}x>4\Rightarrow x>\frac{20}{23}\)
4.
\(4x^2+4x+1-3x+9>4x^2+10\)
\(\Leftrightarrow x>0\)
5.
\(1< \frac{1}{1-x}\Leftrightarrow\frac{1}{1-x}-1>0\Leftrightarrow\frac{x}{1-x}>0\Rightarrow0< x< 1\)
6.
\(\frac{\left(x-5\right)^2\left(x-3\right)}{x+1}\le0\Rightarrow\left[{}\begin{matrix}x=5\\-1< x\le3\end{matrix}\right.\)
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
4 tháng 3 2022
a: \(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)
\(\Leftrightarrow20x^2-12x+15x-5< 20x^2+30x-100\)
=>3x-5<=30x-100
=>30x-100>3x-5
=>27x>95
hay x>95/27
b: \(\Leftrightarrow4\left(5x-2\right)-6\left(2x^2-x\right)< 4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-8-12x^2+6x< 4x-12x^2-15x\)
=>26x-8<-11x
=>37x<8
hay x<8/37
\(\dfrac{2x+1}{x+3}\ge\dfrac{3-5x}{5}+\dfrac{4x+1}{4}\) (ĐK: \(x\ne-3\))
\(\Leftrightarrow\dfrac{20\cdot\left(2x+1\right)}{20\left(x+3\right)}\ge\dfrac{4\left(x+3\right)\left(3-5x\right)}{20\left(x+3\right)}+\dfrac{5\left(4x+1\right)\left(x+3\right)}{20\left(x+3\right)}\)
\(\Leftrightarrow40x+20\ge4\left(3x-5x^2+9-15x\right)+5\left(4x^2+12x+x+3\right)\)
\(\Leftrightarrow40x+20\ge12x-20x^2+36-60x+20x^2+60x+5x+15\)
\(\Leftrightarrow40x+20\ge17x+51\)
\(\Leftrightarrow40x-17x\ge51-20\)
\(\Leftrightarrow23x\ge31\)
\(\Leftrightarrow x\ge\dfrac{31}{23}\left(tm\right)\)
Vậy: \(S=\left\{x\in R|x\le\dfrac{31}{23}\right\}\)