giải BPT :
\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
K
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17 tháng 3 2023
a: =>\(\dfrac{x^2+2x-13-x+1}{x-1}< 0\)
=>\(\dfrac{x^2+x-12}{x-1}< 0\)
=>\(\dfrac{\left(x+4\right)\left(x-3\right)}{x-1}< 0\)
=>1<x<3 hoặc x<-4
b: =>\(\dfrac{3x^2+4x-3x-4}{x-1}< 3\)
=>3x+4<3
=>3x<-1
=>x<-1/3
c: TH1: 2x^2-3x+1>0 và x+2>0
=>(2x-1)(x-1)>0 và x+2>0
=>x>1
TH2: (2x-1)(x-1)<0 và x+2<0
=>x<-2 và 1/2<x<1
=>Loại
4 tháng 3 2022
a: \(\Leftrightarrow15\left(x-1\right)-2\left(7x+3\right)\le10\left(2x+1\right)+6\left(3-2x\right)\)
\(\Leftrightarrow15x-15-14x-6\le20x+10+18-12x\)
=>x-21<=8x+28
=>-7x<=49
hay x>=-7
b: \(\Leftrightarrow20\left(2x+1\right)-15\left(2x^2+3\right)< 10x\left(5-3x\right)-12\left(4x+1\right)\)
\(\Leftrightarrow40x+20-30x^2-45< 50x-30x^2-48x-12\)
=>40x-25<2x-12
=>38x<13
hay x<13/38
4 tháng 3 2022
\(a,\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\\ \Leftrightarrow\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(7x+3\right)}{30}\le\dfrac{10\left(2x+1\right)}{30}+\dfrac{6\left(3-2x\right)}{30}\\ \Leftrightarrow15x-15-14x-6\le20x+10+18-12x\\ \Leftrightarrow x-21\le8x+28\\ \Leftrightarrow7x+49\ge0\\ \Leftrightarrow x\ge-7\)
\(b,\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\\ \Leftrightarrow\dfrac{20\left(2x+1\right)}{-60}-\dfrac{15\left(2x^2+3\right)}{-60}>\dfrac{10x\left(5-3x\right)}{-60}-\dfrac{12\left(4x+1\right)}{-60}\\ \Leftrightarrow40x+20-30x^2-45>50x-30x^2-48x-12\\ \Leftrightarrow38x-13>0\\ \Leftrightarrow x>\dfrac{13}{38}\)
KT
7 tháng 5 2018
\(\dfrac{2x-1}{3}\)+\(\dfrac{x-1}{2}\)\(\le3\)
<=> \(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x-1\right)}{6}\)\(\le\dfrac{18}{6}\)
<=> 4x -2+3x-3\(\le\)18
<=>7x-5\(\le\)18
<=>7x\(\le\)23
<=>x\(\le\)\(\dfrac{23}{7}\)
Vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{23}{7}\)
7 tháng 5 2018
\(\dfrac{2x-1}{3}\)+ \(\dfrac{x-1}{2}\)\(\le\) 3
\(\Leftrightarrow\) \(\dfrac{2.\left(2x-1\right)+3.\left(x-1\right)}{6}\)\(\le\) \(\dfrac{18}{6}\)
\(\Leftrightarrow\) 2.(2x-1)+ 3.( x-1)\(\le\) 18
\(\Leftrightarrow\) 4x- 2+ 3x- 3\(\le\) 18
\(\Leftrightarrow\) 4x+ 3x\(\le\) 18+ 2+ 3
\(\Leftrightarrow\) 7x\(\le\) 23
\(\Leftrightarrow\) x\(\le\) \(\dfrac{23}{7}\)
vậy bpt có no là x\(\le\) \(\dfrac{23}{7}\)
TT
1
HP
20 tháng 2 2021
ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)
\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)
\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)
Vậy ...
NT
1
LA
28 tháng 2 2019
\(\sqrt{2x-1}\ge0\)
\(\Rightarrow BPT\ge0\) khi
\(3-2x-x^2\ge0\)
\(\Leftrightarrow x^2+2x-3\le0\)
\(\Leftrightarrow\left(x+1\right)^2-4\le0\)
\(\Leftrightarrow\left(x+1\right)^2\le4\)
\(\Leftrightarrow x+1\le2\)
\(\Rightarrow x\le1\)
KT
7 tháng 5 2018
\(\dfrac{2x-1}{3}\)-\(\dfrac{x+3}{2}\)\(\le\)1
<=>\(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x+3\right)}{6}\)\(\le\)\(\dfrac{6}{6}\)
=>4x -2 +3x+9\(\le\)6
<=>7x+7\(\le\)6
<=>7x\(\le\)6-7
<=>7x\(\le\)-1
<=>x\(\le\)\(\dfrac{-1}{7}\)
vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{-1}{7}\)
\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
\(\Leftrightarrow6+4x+2>2x-1\)
\(\Leftrightarrow4x-2x>-1-6-2\)
\(\Leftrightarrow x>-\dfrac{9}{2}\)
Vậy S = { x/ x > \(-\dfrac{9}{2}\)}
\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
⇔ \(\dfrac{6+2\left(2x+1\right)}{6}>\dfrac{2x-1}{6}\)
⇔ 6 + 4x + 2 > 2x - 1
⇔ 4x + 8 > 2x - 1
⇔ 2x > - 9
⇔ x > \(\dfrac{-9}{2}\)
KL....