\(\dfrac{x-1}{x+5}=\dfrac{x-2}{x+3}\)
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Tìm x, biếta)dfrac{1}{2}xx-dfrac{7}{3}dfrac{-5}{6}+dfrac{3}{4}xxb)dfrac{4}{5}xx-dfrac{6}{5}dfrac{1}{2}+dfrac{3}{2}xxc)dfrac{2}{5}x(3xx+dfrac{3}{4})1dfrac{1}{5}-dfrac{1}{3}xxd)2x(3xx
+dfrac{3}{4})+dfrac{4}{5}dfrac{1}{2}-2xx
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Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều
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a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
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a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)
1) giải phương trình:a. dfrac{1}{x-5}-dfrac{3}{x^2-6text{x}+5}dfrac{5}{x-1}b. dfrac{x+5}{x-5}-dfrac{x-5}{x+5}dfrac{20}{x^2-25}c.dfrac{x}{left(2text{x}-3right)}+dfrac{x}{2text{x}+2}dfrac{2text{x}}{left(x+1right)left(x-3right)}d.dfrac{x-1}{x+2}-dfrac{x}{x-2}dfrac{5text{x}-2}{4-x^2}e. dfrac{1-6text{x}}{x-2}+dfrac{9text{x}+4}{x+2}dfrac{xleft(3text{x}-2right)+1}{x^2-4}
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1) giải phương trình:
a. \(\dfrac{1}{x-5}-\dfrac{3}{x^2-6\text{x}+5}=\dfrac{5}{x-1}\)
b. \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
c.\(\dfrac{x}{\left(2\text{x}-3\right)}+\dfrac{x}{2\text{x}+2}=\dfrac{2\text{x}}{\left(x+1\right)\left(x-3\right)}\)
d.\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5\text{x}-2}{4-x^2}\)
e. \(\dfrac{1-6\text{x}}{x-2}+\dfrac{9\text{x}+4}{x+2}=\dfrac{x\left(3\text{x}-2\right)+1}{x^2-4}\)
Giải các pt sau:1)dfrac{2x+1}{x^2-4}+dfrac{2}{x+1}dfrac{3}{2-x}2)dfrac{3x+1}{1-3x}+dfrac{3+x}{3-x}23)dfrac{8x-2}{3}1+dfrac{5-2x}{4}4)dfrac{x}{x+1}-dfrac{2x+3}{x}dfrac{-3}{x+1}-dfrac{3}{x}5)dfrac{x+1}{x-1}-dfrac{x-1}{x+1}dfrac{4}{x^2-1}6)dfrac{2x+5}{2x}-dfrac{x}{x+5}0giúp mình với cám ơn
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Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
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12 tháng 3 2022 lúc 14:30
Tìm x biết:a,dfrac{4}{5}+xdfrac{2}{3}b,dfrac{-5}{6}-xdfrac{2}{3}c,dfrac{1}{2}x+dfrac{3}{4}dfrac{-3}{10}d,dfrac{x}{3}-dfrac{1}{2}dfrac{1}{5}e,dfrac{x+3}{15}dfrac{1}{3}h,x+30%x-1,3k,3dfrac{1}{3}x+16dfrac{1}{4}13,25m,dfrac{x-6}{2}dfrac{50}{x-6}n,x-13,424,5-6,7.5,2p,15,7x+3,6x-96,5q,2,5x-11,6-59,1
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Tìm x biết:
\(a,\dfrac{4}{5}+x=\dfrac{2}{3}\)
\(b,\dfrac{-5}{6}-x=\dfrac{2}{3}\)
\(c,\dfrac{1}{2}x+\dfrac{3}{4}=\dfrac{-3}{10}\)
\(d,\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\)
\(e,\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(h,x+30\%x=-1,3\)
\(k,3\dfrac{1}{3}x+16\dfrac{1}{4}=13,25\)
\(m,\dfrac{x-6}{2}=\dfrac{50}{x-6}\)
\(n,x-13,4=24,5-6,7.5,2\)
\(p,15,7x+3,6x=-96,5\)
\(q,2,5x-11,6=-59,1\)
a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
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1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
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4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
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Bài 3:a) dfrac{2x-1}{5}-dfrac{x-2}{3}dfrac{x+7}{15}b) dfrac{x+3}{2}-dfrac{x-1}{3}dfrac{x+5}{6}+1c) dfrac{2left(x+5right)}{3}+dfrac{x+12}{2}-dfrac{5left(x-2right)}{6}dfrac{x}{3}+11d) dfrac{x-4}{5}+dfrac{3x-2}{10}-xdfrac{2x-5}{3}-dfrac{7x+2}{6}e) dfrac{left(2x-3right)left(2x+3right)}{8}dfrac{left(x-4^{ }right)^2}{6}+dfrac{left(x-2right)^2}{3}d) dfrac{7x^2-14x-5}{15}dfrac{left(2x+1right)^2}{5}-dfrac{left(x-1right)^2}{3}e) dfrac{left(7x+1right)left(x-2right)}{10}+dfrac{2}{5}dfrac{left(x-2right)^2}{5...
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Bài 3:
a) \(\dfrac{2x-1}{5}\)-\(\dfrac{x-2}{3}\)
=\(\dfrac{x+7}{15}\)
b) \(\dfrac{x+3}{2}\)-\(\dfrac{x-1}{3}\)
=\(\dfrac{x+5}{6}\)+1
c) \(\dfrac{2\left(x+5\right)}{3}\)+\(\dfrac{x+12}{2}\)
-\(\dfrac{5\left(x-2\right)}{6}\)=\(\dfrac{x}{3}\)+11
d) \(\dfrac{x-4}{5}\)+\(\dfrac{3x-2}{10}\)-x
=\(\dfrac{2x-5}{3}\)-\(\dfrac{7x+2}{6}\)
e) \(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\)
=\(\dfrac{\left(x-4^{ }\right)^2}{6}\)+\(\dfrac{\left(x-2\right)^2}{3}\)
d) \(\dfrac{7x^2-14x-5}{15}\)
=\(\dfrac{\left(2x+1\right)^2}{5}\)-\(\dfrac{\left(x-1\right)^2}{3}\)
e) \(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}\)+\(\dfrac{2}{5}\)
=\(\dfrac{\left(x-2\right)^2}{5}\)+\(\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)
a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
\(\Leftrightarrow6x-3-5x+10-x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
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tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
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Tìm x, biết:
a) dfrac{3}{7}x - dfrac{2}{3}x dfrac{10}{21}
b) dfrac{7}{35} : (x - dfrac{1}{3}) -dfrac{2}{25}
c) 3.(x - dfrac{1}{2}) - 5. (x + dfrac{3}{5}) -x + dfrac{1}{5}
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Tìm x, biết:
a) \(\dfrac{3}{7}\)x - \(\dfrac{2}{3}\)x = \(\dfrac{10}{21}\)
b) \(\dfrac{7}{35}\) : (x - \(\dfrac{1}{3}\)) = \(-\dfrac{2}{25}\)
c) 3.(x - \(\dfrac{1}{2}\)) - 5. (x + \(\dfrac{3}{5}\)) = -x + \(\dfrac{1}{5}\)
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
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\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
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a,73x−32x=2110⇒x(73−32)=2110⇒x.−215=2110⇒x=−2b,357:(x−31)=−252⇒51:(x−31)=−252⇒x−31=−25⇒x=−613c,3.(x−21)−5.(x+53)=−x+51⇒3x−23−5x+5=−x+51
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1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
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1)\(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)
\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)
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4: Ta có: \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow40x-20+45x-30=48x-36\)
\(\Leftrightarrow37x=14\)
hay \(x=\dfrac{14}{37}\)
5: Ta có: \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
\(\Leftrightarrow2x-6-3x-6=x+4-9\)
\(\Leftrightarrow-x-x=-5-12=-17\)
hay \(x=\dfrac{17}{2}\)
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