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\(\left(\dfrac{103}{8}-\dfrac{193}{18}\right):x-\dfrac{40}{33}:\dfrac{8}{11}=\dfrac{5}{3}\)
\(\dfrac{155}{72}:x-\dfrac{5}{3}=\dfrac{5}{3}\)
\(\dfrac{155}{72}:x=\dfrac{5}{3}+\dfrac{5}{3}=\dfrac{10}{3}\)
\(x=\dfrac{155}{72}:\dfrac{10}{3}=\dfrac{31}{48}\)
vaayj.....
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
\(\dfrac{7}{9}+\dfrac{1}{3} < x < \dfrac{43}{8}+\dfrac{1}{10}\)
\(\dfrac{10}{9} < x < \dfrac{219}{40}\)
Mà \(x \in N\)
\(=>x=\){`2;3;4;5`}
\(\dfrac{7}{9}+\dfrac{1}{3}< x< \dfrac{43}{8}+\dfrac{1}{10}\)
\(\dfrac{10}{9}< x< \dfrac{219}{40}\)
Mà \(x\inℕ\)
\(\Rightarrow\dfrac{10}{9}< 2\le x\le5< \dfrac{219}{40}\)
\(\Rightarrow2\le x\le5\)
\(\Rightarrow x\in\left\{2;3;4;5\right\}\)
Vậy: \(x\in\left\{2;3;4;5\right\}\)
Bài 1 :
\(=\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}-\dfrac{3}{8}-\dfrac{5}{8}=0-1=-1\)
Bài 2 :
\(\Rightarrow3+x=8\Leftrightarrow x=5\)
Bài 3 :
\(\Leftrightarrow x-\dfrac{5}{11}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{35}{44}\)
Bài 4 :
Trong 2 ngày An đọc được số quyên phần quyên sách
\(\dfrac{1}{11}+\dfrac{8}{11}=\dfrac{9}{11}\)( quyển sách )
đs : 9/11 quyển sách
a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)
hay \(x=\dfrac{6}{25}\)
Vậy: \(x=\dfrac{6}{25}\)
b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)
hay \(x=\dfrac{5}{99}\)
Vậy: \(x=\dfrac{5}{99}\)
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, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)
`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`
`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`
`<=>(x+20)(1/12+1/11+1/10)=0`
Vì `1/12+1/11+1/10 ≠ 0`
`=>x+20=0`
`=>x=0-20`
`=>x=-20`