tìm x :2/x-2/+13=15
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hộ mik nha
a)\(2.x-\frac{3}{2}=\frac{1}{2}\)
\(2x=\frac{4}{2}\)
\(x=\frac{4}{2}:2\)
\(x=1\)
Vậy x=1
b)\(\frac{8}{13}.x+\frac{5}{13}.x=\frac{1}{10}\)
\(x.\left(\frac{8}{13}+\frac{5}{13}\right)=\frac{1}{10}\)\
\(x.1=\frac{1}{10}\)
\(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c)\(\frac{2}{15}+\frac{7}{15}.x=\frac{1}{15}\)
\(\frac{7}{15}x=\frac{-1}{15}\)
x=\(\frac{-1}{7}\)
Vậy \(x=\frac{-1}{7}\)
\(\frac{3}{2}\cdot x-\frac{3}{2}=\frac{1}{2}\) \(\frac{8}{13}\cdot x+\frac{5}{13}\cdot x=\frac{1}{10}\)
\(\frac{3}{2}\cdot x=\frac{3}{2}+\frac{1}{2}=2\) \(\left[\frac{8}{13}+\frac{5}{13}\right]\cdot x=\frac{1}{10}\)
\(x=2:\frac{3}{2}\) \(1\cdot x=\frac{1}{10}\)
x=\(\frac{4}{3}\) \(x=\frac{1}{10}:1\)
Vậy x=\(\frac{4}{3}\) \(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c,\(\frac{2}{5}+\frac{7}{15}\cdot x=\frac{1}{15}\)
\(\frac{7}{15}\cdot x=\frac{1}{15}-\frac{2}{5}=\frac{-1}{3}\)
\(x=\frac{-1}{3}:\frac{7}{15}\)
\(x=\frac{-5}{7}\)
Vậy x=\(\frac{-5}{7}\)
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+\frac{2}{15\times17}+\frac{2}{17\times19}+\frac{2}{19\times21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\times462-x=19\)
\(\frac{10}{231}\times462-x=19\)
\(20-x=19\)
\(\Rightarrow x=20-19=1\)
bạn viết sai đề bài nhé
(x+2)/11+(x+2)/12+(x+2)/13=(x+2)/14+(x+2)15
<=> (x+2)/11+(x+2)/12+(x+2)/13 - (x+2)/14 - (x+2)/15 = 0
<=> (x+2)(1/11+1/12+1/13 - 1/14 - 1/15 ) = 0
vì: (1/11+1/12+1/13 - 1/14 - 1/15 ) khác 0 nên x-2 = 0 => x=2
`(2/(11.13)+2/(13.15)+....+2/(19.21)).462-[0,04/(x+1,05):0,12=19`
`=>(1/11-1/13+1/13-1/15+....+1/19-1/21).462-(2/(50(x+1,05)).25/3=19`
`=>(1/11-1/21).462-1/(3(x+1,05))=19`
`=>10/231. 462-1/(3x+3,15)=19`
`=>20-1/(3x+3,15)=19
`=>1/(3x+3,15)=1`
`=>3x+3,15=1`
`=>3x=-2,15`
`=>x=-43/60`
Vậy `x=-43/60`
Bạn xem lại đề nha \(\dfrac{x+2}{15}\)
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
=> \(\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
=> x=-2 do \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\)\(\ne0\)
\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
\(\Rightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)
\(\Rightarrow\left(x+2\right)\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\Rightarrow x+2=0\left(\text{Vì }\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\right)\)
=>x=-2
Vậy x=-2
\(2.\left|x-2\right|+13=15\)
\(\Rightarrow2.\left|x-2\right|=2\)
\(\Rightarrow\left|x-2\right|=1\)
Nếu \(x-2\ge0\) thì \(x\ge2\)
\(\Rightarrow x-2=1\Rightarrow x=3\) (nhận)
Nếu \(x-2< 0\) thì x < 2
\(\Rightarrow-x+2=1\Rightarrow-x=-1\Rightarrow x=1\) (nhận)
Vậy x = {3;1}
ko hiểu đề bài
ak tớ hiểu rồi
x=3 hoặc x= 1