b)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà. b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\) =>\(\dfrac{-3}{5}.x=1\) =>\(x=1:\dfrac{-3}{5}\) =>\(x=\dfrac{-5}{3}\) Vậy \(x=\dfrac{-5}{3}\) a) \(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\cdot\left(2x-1\right)\\
\dfrac{-2}{3}x-\left(\dfrac{-1}{6}\right)=\dfrac{2}{3}x-\dfrac{1}{3}\\
\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\
\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\\
-x\cdot\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{-2}{6}-\dfrac{1}{6}\\
\dfrac{-4}{3}x=\dfrac{-1}{2}\\
x=\dfrac{-1}{2}:\dfrac{-4}{3}\\
x=\dfrac{-1}{2}\cdot\dfrac{-3}{4}\\
x=\dfrac{3}{8}\) a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\) <=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\) <=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\) <=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\) <=>\(4x-17=0\) <=>\(4x=17\)<=>\(x=\dfrac{17}{4}\) a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\) \(\dfrac{3}{10}-x=\dfrac{7}{10}\) x = \(\dfrac{3}{10}-\dfrac{7}{10}\) x=\(\dfrac{-4}{10}\) b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\) \(\dfrac{-5}{9}+x=\dfrac{-59}{9}\) \(x=\dfrac{-59}{9}-\dfrac{-5}{9}\) \(x=\dfrac{-64}{9}\) c)=>2.18=(x-3).(x-3) =>36=(x-3)\(^2\) =>6\(^2\)=(x-3)\(^2\) 6= x-3 x=6+3=9 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 Bài 1: \(a,\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\) \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{24+2-3}{12}\right)=\dfrac{7}{46}\) \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\) \(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{46}.\dfrac{23}{12}\) \(x+\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{7}{24}\) \(x+\dfrac{1}{4}=\dfrac{7}{24}+\dfrac{1}{3}\) \(x+\dfrac{1}{4}=\dfrac{5}{8}\) \(x=\dfrac{5}{8}-\dfrac{1}{4}=\dfrac{3}{8}\) Vậy \(x=\dfrac{3}{8}\) \(b,\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{7}{10}\) \(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\) \(\left(\dfrac{13}{21}+x\right).\dfrac{7}{12}=\dfrac{1}{6}\) \(\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\) \(\dfrac{13}{21}+x=\dfrac{2}{7}\) \(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{1}{3}\) Vậy \(x=-\dfrac{1}{3}\) Bài 2: \(a,\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\) \(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(\dfrac{121}{12}-\dfrac{19}{2}\right)\) \(=\dfrac{77}{18}:\dfrac{7}{12}\) \(=\dfrac{22}{3}\) \(b,1\dfrac{5}{18}-\dfrac{5}{18}.\left(\dfrac{1}{15}+1\dfrac{1}{12}\right)\) \(=\dfrac{23}{18}-\dfrac{5}{18}.\dfrac{69}{60}\) \(=\dfrac{23}{18}-\dfrac{23}{72}\) \(=\dfrac{23}{24}\) \(c,-\dfrac{1}{7}.\left(9\dfrac{1}{2}-8,75\right):\dfrac{2}{7}+0,625:1\dfrac{2}{3}\) \(=\dfrac{-1}{7}.\dfrac{3}{4}:\dfrac{2}{7}+\dfrac{5}{8}:\dfrac{5}{3}\) \(=-\dfrac{3}{8}+\dfrac{5}{8}:\dfrac{5}{3}\) \(=-\dfrac{3}{8}+\dfrac{3}{8}\) \(=\dfrac{0}{8}=0\) Chúc bạn học tốt ukm bn có thể giải cho mik mấy bài mà mik vừa đăng đc ko mik đang cần gấp a, (\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)).10 - x = 0 Bài 1: a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\) =>x+4/15=8/5 hoặc x+4/15=-8/5 =>x=4/3 hoặc x=-28/15 b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\) c: \(\Leftrightarrow\left|x-1\right|-1=1\) =>|x-1|=2 =>x-1=2 hoặc x-1=-2 =>x=3 hoặc x=-1 Bài 2: b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\) Bài 3: a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\) Dấu '=' xảy ra khi x=-15/19 b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\) Dấu '=' xảy ra khi x=4/7
K
Khách
TH
Thanh Hoàng Thanh
11 tháng 3 2022
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<=> \(\dfrac{5}{6}.10-x=0\)
<=> \(\dfrac{25}{3}-x=0\)
<=> x = \(\dfrac{25}{3}\) (thỏa mãn)
@Hoàng Mạnh Quân
29 tháng 7 2022
