Tìm x
X - 1\(\dfrac{1}{2}\) = 2 \(\dfrac{3}{5}\)
1/3 + 12/5 : x = 3\(\dfrac{2}{3}\)
3384 : 94 : x = 9
Tìm x :
a, \(\dfrac{11}{12}\) - (\(\dfrac{2}{5}\) +x) = \(\dfrac{2}{3}\) b, x+\(\dfrac{2}{3}\) =\(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\) c, x-[\(\dfrac{17}{2}\) -(\(\dfrac{-3}{7}\) + \(\dfrac{5}{3}\) )]=\(\dfrac{-1}{3}\)
d,\(\dfrac{9}{2}\) - [\(\dfrac{2}{3}\) - (x+\(\dfrac{7}{4}\) )] =\(\dfrac{-5}{4}\)
Mọi người giúp mình nha
b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)
\(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)
\(x\) = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{1}{10}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x\) = \(\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x\) = \(\dfrac{1}{4}\)
x = \(\dfrac{2}{3}-\dfrac{1}{4}\)
\(x=\dfrac{5}{12}\)
d, \(\dfrac{9}{2}\) - ( \(\dfrac{2}{3}\) - (\(x\) + \(\dfrac{7}{4}\))] = \(\dfrac{-5}{4}\)
\(\dfrac{9}{2}\) - \(\dfrac{2}{3}\) + \(x\) + \(\dfrac{7}{4}\) = \(-\dfrac{5}{4}\)
\(\dfrac{23}{6}\) + \(x\) = - \(\dfrac{5}{4}\) - \(\dfrac{7}{4}\)
\(x\) = - 3 - \(\dfrac{23}{6}\)
\(x\) = - \(\dfrac{41}{6}\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
Tìm x:
\(\dfrac{3}{5}\) : x = \(\dfrac{1}{3}+\dfrac{1}{2}\) x : \(\dfrac{7}{12}\) = 2 x - \(\dfrac{3}{2}\) = \(\dfrac{11}{4}-\dfrac{5}{4}\) x + \(\dfrac{5}{4}\) = \(\dfrac{3}{2}+\dfrac{7}{12}\)
`3/5 : x =1/3 +1/2`
` 3/5 : x= 2/6 +3/6`
` 3/5 : x= 5/6`
` x= 3/5 : 5/6`
` x= 3/5 xx 6/5`
` x= 18/25`
__
`x: 7/15 =2`
` x= 2xx 7/15`
` x= 14/15`
__
`x-3/2=11/4-5/4`
`x-3/2= 6/4`
`x= 3/2 +3/2`
`x= 6/2`
`x=3`
__
`x+5/4 = 3/2+7/12`
`x+5/4 = 18/12+7/12`
`x+5/4 = 25/12`
`x= 25/12-5/4`
`x= 25/12- 15/12`
`x= 10/12`
`x= 5/6`
1) \(\dfrac{5x-2}{3}\)= \(\dfrac{5-3x}{2}\)
2) \(\dfrac{x+4}{5}\) - x + 4 = \(\dfrac{x}{3}\) - \(\dfrac{x-2}{2}\)
3) \(\dfrac{10x+3}{12}\)= 1 + \(\dfrac{6+8x}{9}\)
4) \(\dfrac{x+1}{3}\)- \(\dfrac{x-2}{6}\) = \(\dfrac{2x-1}{2}\)
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
2) Ta có: ⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+144+5x−30=0⇔−24x+144+5x−30=0
⇔−19x+114=0⇔−19x+114=0
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: x=6
3) Ta có: ⇔3(10x+3)36=3636+4(6+8x)36⇔3(10x+3)36=3636+4(6+8x)36
⇔30x+9=36+24+32x⇔30x+9=36+24+32x
⇔30x+9−60−32x=0⇔30x+9−60−32x=0
⇔−2x−51=0⇔−2x−51=0
⇔−2x=51⇔−2x=51
hay x=−512x=−512
4) Ta có: ⇔2(x+1)6−x−26=3(2x−1)6⇔2(x+1)6−x−26=3(2x−1)6
⇔2x+2−x+2=6x−3⇔2x+2−x+2=6x−3
⇔x+4−6x+3=0⇔x+4−6x+3=0
⇔−5x+7=0⇔−5x+7=0
⇔−5x=−7⇔−5x=−7
hay x=75
g) \(3-\dfrac{2}{2x-3}=\dfrac{2}{5}=\dfrac{2}{9-6x}-\dfrac{3}{2}\)
h) \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
i) \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)
k) \(\dfrac{13}{x-1}+\dfrac{5}{2x-2}-\dfrac{6}{3x-3}\)
m) \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)
n) \(\left(\dfrac{3}{2}-\dfrac{5}{11}-\dfrac{3}{13}\right)\left(2x-2\right)=\left(-\dfrac{3}{4}+\dfrac{5}{22}+\dfrac{3}{26}\right)\)
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)
\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)
\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)
Vậy ...
i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)
\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)
Vậy ...
h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)
\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)
\(\Leftrightarrow12x^2-24-2x=0\)
\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)
m) Ta có: \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{19}{10}:x=2\)
hay \(x=\dfrac{19}{20}\)
Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)
giúp mình với ạ
Bài 1: giải các PT:
a, \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
b, \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
c, \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
d, \(\dfrac{x+\dfrac{2\left(3-x\right)}{5}}{14}-\dfrac{5x-4\left(x-1\right)}{24}=\dfrac{7x+2+\dfrac{9-3x}{5}}{12}+\dfrac{2}{3}\)
\(e,\dfrac{x-\dfrac{3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}}{ }\)
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 ................................
haizzz bệnh lười lại lên cơn r
tìm x , biết:
a) \(x\) : \(4\dfrac{1}{3}\) = -2,5 b) \(\dfrac{3}{5}x\) + \(\dfrac{1}{4}\) = \(\dfrac{1}{10}\)
c) \(2\dfrac{7}{9}\) \(-\) \(\dfrac{12}{13}x\) = \(\dfrac{7}{9}\) d)\(\dfrac{-2}{3}-\dfrac{1}{3}\)\(\left(2x-5\right)=\dfrac{3}{2}\)
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
tìm x
\(\dfrac{3-x}{5-x}=\dfrac{6}{11}\) \(\left(1\dfrac{1}{3}-25\%.x-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)
\(\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)-\dfrac{1}{3}.\left(2x-3\right)=x\)
\(2.\left(\dfrac{1}{2}-x\right)-3\left(x-\dfrac{1}{3}\right)=\dfrac{7}{2}\)
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10