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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
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 \(x=\dfrac{-5}{3}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
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
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
9:
\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=\left(5-2-8\right)+\left(\dfrac{1}{5}+\dfrac{73}{35}-\dfrac{2}{7}\right)+\left(-\dfrac{2}{9}+\dfrac{1}{18}-\dfrac{5}{6}\right)+\dfrac{1}{23}\)
\(=\left(-5\right)+\dfrac{7+73-10}{35}+\dfrac{-4+1-15}{18}+\dfrac{1}{23}\)
\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)
10: \(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}-\dfrac{2}{9}-\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)
=1/64
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a/ \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{20}\)
\(\Leftrightarrow x=\dfrac{29}{10}\)
Vậy ...
b/ \(\left(4x-3\right)\left(\dfrac{5}{4}x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{8}{5}\end{matrix}\right.\)
Vậy .....
c/ \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\Leftrightarrow\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=-\dfrac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=-\dfrac{19}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{38}{21}\end{matrix}\right.\)
Vậy ......
d/ \(\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)
\(\Leftrightarrow\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(\Leftrightarrow\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{9}{10}\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a. \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{4}\)
\(\dfrac{3}{6}x=\dfrac{5}{4}+\dfrac{1}{5}\)
\(\dfrac{3}{6}x=\dfrac{29}{20}\)
\(x=\dfrac{29}{20}:\dfrac{3}{6}\)
\(x=\dfrac{29}{10}\)
Vậy...
b. \(\left(4x-3\right).\left(\dfrac{5}{4}x+2\right)=0\)
\(\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-8}{5}\end{matrix}\right.\)
Vậy ...
c. \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=1,5+\dfrac{3}{4}\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{-9}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=\dfrac{-19}{12}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{-38}{21}\end{matrix}\right.\)
Vậy...
\(a,\left(x-\dfrac{5}{8}\right).\dfrac{5}{8}=-\dfrac{15}{36}\)
\(\left(x-\dfrac{5}{8}\right)=-\dfrac{15}{36}\div\dfrac{5}{8}\)
\(x-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+\dfrac{5}{8}\)
\(x=-\dfrac{1}{24}\)
\(b,\left(x-\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}\)
\(a,\left(x-\dfrac{5}{8}\right)\cdot\dfrac{8}{18}=-\dfrac{15}{16}\\ x-\dfrac{5}{8}=-\dfrac{15}{36}:\dfrac{8}{18}\\ x-\dfrac{5}{8}=-\dfrac{15}{16}\\ x=-\dfrac{15}{16}+\dfrac{5}{8}\\ x=-\dfrac{15}{16}+\dfrac{10}{16}\\ x=-\dfrac{5}{16}\\ b,x-\dfrac{1}{3}=\dfrac{5}{6}\\ x=\dfrac{5}{6}+\dfrac{1}{3}\\ x=\dfrac{5}{6}+\dfrac{2}{6}\\ x=\dfrac{7}{6}\)