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a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)
\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(x=\dfrac{-3}{4}\)
b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)
\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(x=-\dfrac{2}{3}\)
c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)
\(x=\dfrac{1}{5}\)
d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)
\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)
\(\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(x=\dfrac{16}{15}\)
#YVA6
\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)
\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)
\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)
\(\Leftrightarrow x=-\dfrac{3}{4}\)
\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)
\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)
\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)
\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{16}{15}\)
Giải:
a)-3/10-(-1/5)+x)=-3/2
-1/5+x =-3/10-(-3/2)
-1/5+x =6/5
x =6/5-(-1/5)
x =7/5
b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)
x-3/4+16/5 =1/2
x-3/4 =1/2-16/5
x =-27/10
x =-27/10+3/4
x =-39/20
c)x-3/x+5=4/3
=>(x-3).3=4.(x+5)
3x-9 =4x+20
3x-4x =20+9
-1x =29
x =-29
Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!
Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2
Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2
Chúc bạn học tốt!
\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)
`<=>((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`
\(TH1:x\ge0\)
\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)
\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)
\(TH2:x< 0\)
\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)
\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)
Vậy ...
Giải:
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(TH1:x\ge0\)
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)
\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\)
\(x.\dfrac{1}{6}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:\dfrac{1}{6}\)
\(x=5\)
\(TH2:x\le0\)
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\)
\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\)
\(x.\dfrac{-11}{6}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:\dfrac{-11}{6}\)
\(x=\dfrac{-5}{11}\)
Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
\(\left(\dfrac{103}{8}-\dfrac{193}{18}\right):x-\dfrac{40}{33}:\dfrac{8}{11}=\dfrac{5}{3}\)
\(\dfrac{155}{72}:x-\dfrac{5}{3}=\dfrac{5}{3}\)
\(\dfrac{155}{72}:x=\dfrac{5}{3}+\dfrac{5}{3}=\dfrac{10}{3}\)
\(x=\dfrac{155}{72}:\dfrac{10}{3}=\dfrac{31}{48}\)
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) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
\(a,\dfrac{x}{8}=\dfrac{7}{-2}\\ \Rightarrow x=-28\\ b,\dfrac{1-2x}{6}=\dfrac{-1}{2}\\ \Leftrightarrow2-4x=-6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\\ c,\dfrac{x+2}{3}=\dfrac{x+3}{4}\\ \Leftrightarrow4x+8=3x+9\\ \Leftrightarrow x=1\\ d,\dfrac{10}{2-x}=2\\ \Leftrightarrow4-2x=10\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)
\(\dfrac{7}{9}+\dfrac{1}{3} < x < \dfrac{43}{8}+\dfrac{1}{10}\)
\(\dfrac{10}{9} < x < \dfrac{219}{40}\)
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: \(x\in\left\{2;3;4;5\right\}\)
\(\dfrac{x}{-8}=\dfrac{-15}{10}\\ \Rightarrow10x=\left(-8\right)\left(-15\right)=120\\ \Rightarrow x=12\)
Vậy x = 12
x=\(\dfrac{-15}{10}.\left(-8\right)\)=12