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a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
\(a,\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=-3+3\)
\(\Leftrightarrow\dfrac{1+x+2017}{2017}+\dfrac{2+x+2016}{2016}+\dfrac{3+x+2015}{2015}=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b,\(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{2x+4}{5}}{15}=\dfrac{\dfrac{11x-3}{2}}{5}-\dfrac{5x-5}{5}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-\dfrac{10x-10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3-10x+10}{10}\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{x+7}{10}\)
\(\Leftrightarrow10\left(2x+4\right)=75\left(x+7\right)\)
\(\Leftrightarrow20x+40=75x+525\)
\(\Leftrightarrow20x-75x=525-40\)
\(\Leftrightarrow-55x=485\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
a) \(\dfrac{1+x}{2017}+\dfrac{2+x}{2016}+\dfrac{3+x}{2015}=-3\)
\(\Leftrightarrow\dfrac{1+x}{2017}+1+\dfrac{2+x}{2016}+1+\dfrac{3+x}{2015}+1=0\)
\(\Leftrightarrow\dfrac{x+2018}{2017}+\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}+\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(\dfrac{x-\dfrac{3x-4}{5}}{15}=\dfrac{5x-\dfrac{3-x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{\dfrac{5x-3x+4}{5}}{15}=\dfrac{\dfrac{10x-3+x}{2}}{5}-x+1\)
\(\Leftrightarrow\dfrac{2x+4}{75}=\dfrac{11x-3}{10}-x+1\)
\(\Leftrightarrow\dfrac{4x+8}{150}=\dfrac{165x-45}{150}-\dfrac{150x-150}{150}\)
\(\Leftrightarrow4x+8=165x-45-150x+150\)
\(\Leftrightarrow4x-165x+150x=-45+150-8\)
\(\Leftrightarrow-11x=97\)
\(\Leftrightarrow x=-\dfrac{97}{11}\)
\(S=\left\{-\dfrac{97}{11}\right\}\)
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
bạn nên bổ sung chữ "bất"
1)
\(x-\dfrac{x-1}{3}+\dfrac{x+2}{6}>\dfrac{2x}{5}+5\\ \Leftrightarrow x-\dfrac{x-1}{3}+\dfrac{x+2}{6}-\dfrac{2x}{5}-5>0\\ \Leftrightarrow\dfrac{30x-10\left(x-1\right)+5\left(x+2\right)-2x\cdot6-5\cdot30}{30}>0\\ \Leftrightarrow30x-10x+10+5x+10-12x-150>0\\ \Leftrightarrow30x-10x=5x-12x>-10-10+150\\ \Leftrightarrow13x>130\\ \Leftrightarrow13x\cdot\dfrac{1}{13}>130\cdot\dfrac{1}{13}\\ \Leftrightarrow x>10\)
Vậy tập ngiệm của bât hương trình là {x/x>10}
mình mới học đến đây nên cách giải còn dài, thông cảm nha
2)
\(\dfrac{2x+6}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{x+3}{3}-\dfrac{x-2}{9}-1< 0\\ \Leftrightarrow\dfrac{3\left(x+3\right)-x+2-9}{9}< 0\\ \Leftrightarrow3x+9-x+2-9< 0\\ \Leftrightarrow3x-x< -9+9-2\\ \Leftrightarrow2x< -2\\ \Leftrightarrow2x\cdot\dfrac{1}{2}< -2\cdot\dfrac{1}{2}\Leftrightarrow x< -1\)
Vậy tập nghiệm của bất phương trình là {x/x<-1}
a) \(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)
\(\Leftrightarrow84x+63-90x+30=175x+140+315\)
\(\Leftrightarrow84x-90x-175x=140+315-63-30\)
\(\Leftrightarrow-181x=362\)
\(\Leftrightarrow x=-\dfrac{362}{181}=-2\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-2\right\}\)
b) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
\(\Leftrightarrow6x+24-30x-120=10x-15x+30\)
\(\Leftrightarrow6x-30x-10x+15x=30-24+120\)
\(\Leftrightarrow-19x=126\)
\(\Leftrightarrow x=-\dfrac{126}{19}\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-\dfrac{126}{19}\right\}\)
c) \(\dfrac{x+2}{3}+\dfrac{3\left(2x-1\right)}{4}-\dfrac{5x-3}{6}=x+\dfrac{5}{12}\)
\(\Leftrightarrow4\left(x+2\right)+9\left(2x-1\right)-2\left(5x-3\right)=12x+5\)
\(\Leftrightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow4x+18x-10x-12x=5-8+9-6\)
\(\Leftrightarrow0x=0\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{R\right\}\)
_Chúc bạn học tốt_
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
a) 4( x - 2 ) - 3 ( x - 3 ) = 1
4x - 8 - 3x + 9 =1
x = 0
a)\(\dfrac{x-2}{3}-\dfrac{x-3}{4}=1\Leftrightarrow\dfrac{4x-8-3x+9}{12}=1\) ⇔x+1=12⇔x=11 Vậy phương trình đã cho có tập nghiệm S=\(\left\{11\right\}\) b)\(\dfrac{x-1}{2015}+\dfrac{x-2}{2014}+\dfrac{x-5}{2011}+\dfrac{x+1}{2017}=4\) \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+\left(\dfrac{x-5}{2011}-1\right)+\left(\dfrac{x+1}{2017}-1\right)=4-4\) \(\Leftrightarrow\dfrac{x-1-2015}{2015}+\dfrac{x-2-2014}{2014}+\dfrac{x-5-2011}{2011}+\dfrac{x+1-2017}{2017}=0\) \(\Leftrightarrow\dfrac{x-2016}{2015}+\dfrac{x-2016}{2014}+\dfrac{x-2016}{2011}+\dfrac{x-2016}{2017}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x-2016=0\) (vì \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2011}+\dfrac{1}{2017}\ne0\) )
⇔x=2016
Vậy phương trình đã cho có tập nghiệm S=\(\left\{2016\right\}\)
c)3(x-1)-5(x+4)+6(2-x)=7 ⇔3x-3-5x-20+12-6x=7⇔3x-5x-6x=7-12+20+3⇔-8x=18⇔\(x=\dfrac{-9}{4}\)
Vậy phương trình đã cho có tập nghiệm S=\(\left\{\dfrac{-9}{4}\right\}\)