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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\) \(-\dfrac{5}{6}x=\dfrac{5}{12}\) \(x=-\dfrac{1}{2}\) b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\) \(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\) \(3x-3.7=-\dfrac{19}{2}\) \(3x=-5.8\) \(x=-\dfrac{29}{15}\) c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\) \(2+\dfrac{3}{4}x=\dfrac{21}{8}\) \(\dfrac{3}{4}x=\dfrac{5}{8}\) \(x=\dfrac{5}{6}\) d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\) \(-\dfrac{2}{3}x=\dfrac{1}{10}\) \(x=-\dfrac{3}{20}\) Bài 1: a) \(\dfrac{x-1}{9}=\dfrac{8}{3}\\
\Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\
\Leftrightarrow x-1=24\\
x=24+1\\
x=25\) b) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\
\dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\
\dfrac{3x}{7}+1=\dfrac{1}{2}\\
\dfrac{3x}{7}=\dfrac{1}{2}-1\\
\dfrac{3x}{7}=\dfrac{-1}{2}\\
3x=\dfrac{-1}{2}\cdot7\\
3x=\dfrac{-7}{2}\\
x=\dfrac{-7}{2}:3\\
x=\dfrac{-7}{6}\) c) \(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\
x+\dfrac{7}{12}=\dfrac{5}{6}\\
x=\dfrac{5}{6}-\dfrac{7}{12}\\
x=\dfrac{1}{4}\) d) \(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\
\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\
x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\
\dfrac{-1}{6}x=\dfrac{7}{12}\\
x=\dfrac{7}{12}:\dfrac{-1}{6}\\
x=\dfrac{-7}{2}\) e) \(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\
\dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\
\dfrac{277}{690}-x=\dfrac{7}{46}\\
x=\dfrac{277}{690}-\dfrac{7}{46}\\
x=\dfrac{86}{345}\) f) \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\
\left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\
x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\
x-\dfrac{1}{12}=\dfrac{7}{24}\\
x=\dfrac{7}{24}+\dfrac{1}{12}\\
x=\dfrac{3}{8}\) g) \(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\
\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\
\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\
\dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\
\dfrac{13}{21}+x=\dfrac{2}{7}\\
x=\dfrac{2}{7}-\dfrac{13}{21}\\
x=\dfrac{-1}{3}\) h) \(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\
2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\
2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\
\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\
\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\
\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\
\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\
\dfrac{1}{2}x=\dfrac{29}{24}\\
x=\dfrac{29}{24}:\dfrac{1}{2}\\
x=\dfrac{29}{12}\\
\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\) i) \(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\
3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\
3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\
\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\
\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\
\left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\
\Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\
3x=\dfrac{-1}{3}+\dfrac{1}{2}\\
3x=\dfrac{1}{6}\\
x=\dfrac{1}{6}:3\\
x=\dfrac{1}{18}\) tìm x a) c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\) \(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\) \(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\) \(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\) \(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\) \(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\) \(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\) \(\Rightarrow x=-36\) Vậy ... a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\) \(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\) \(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\) \(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\) \(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\) Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\) Chúc bạn học tốt!!! a) \(\left(2x-3\right)\left(6-2x\right)=0\) \(\circledast\)TH1: \(2x-3=0\\
2x=0+3\\
2x=3\\
x=\dfrac{3}{2}\) \(\circledast\)TH2: \(6-2x=0\\
2x=6-0\\
2x=6\\
x=\dfrac{6}{2}=3\) Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\). b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\) \(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\) \(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\) \(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\) \(-\dfrac{11}{15}=-x\left(x-1\right)\) \(\Rightarrow x=1.491631652\) Vậy \(x=1.491631652\) c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\) \(\circledast\)TH1: \(3x-1=0\\
3x=0+1\\
3x=1\\
x=\dfrac{1}{3}\) \(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\
-\dfrac{1}{2}x=0-5\\
-\dfrac{1}{2}x=-5\\
x=-5:-\dfrac{1}{2}\\
x=10\) Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\). d) \(\dfrac{x}{5}=\dfrac{2}{3}\\
x=\dfrac{5\cdot2}{3}\\
x=\dfrac{10}{3}\) Vậy \(x=\dfrac{10}{3}\). e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\
\) \(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\) \(\dfrac{x}{3}=\dfrac{7}{10}\) \(x=\dfrac{3\cdot7}{10}\) \(x=\dfrac{21}{10}\) Vậy \(x=\dfrac{21}{10}\). f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\) \(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\) \(\dfrac{x}{5}=\dfrac{11}{10}\) \(x=\dfrac{5\cdot11}{10}\) \(x=\dfrac{55}{10}=\dfrac{11}{2}\) Vậy \(x=\dfrac{11}{2}\). g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\
x+3=\dfrac{15}{3}=5\\
x=5-3\\
x=2\) Vậy \(x=2\). h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\
x-12=\dfrac{4}{2}=2\\
x=2+12\\
x=14\) Vậy \(x=14\). 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 a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\) \(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\) \(-\dfrac{2}{3}x=\dfrac{1}{10}\) x=\(\dfrac{1}{10}:-\dfrac{2}{3}\) \(x=-\dfrac{3}{20}\) Vậy \(x=-\dfrac{3}{20}\). b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\) \(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\) \(\dfrac{2}{3}:x=-\dfrac{22}{3}\) \(x=\dfrac{2}{3}:-\dfrac{22}{3}\) \(x=-\dfrac{1}{11}\) Vậy \(x=-\dfrac{1}{11}\). c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\) \(60\%x=\dfrac{19}{9}\) \(\dfrac{3}{5}x=\dfrac{19}{9}\) \(x=\dfrac{19}{9}:\dfrac{3}{5}\) \(x=\dfrac{95}{27}\) Vậy \(x=\dfrac{95}{27}\). d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\) \(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\) \(\dfrac{2}{3}-x=\dfrac{3}{20}\) \(x=\dfrac{2}{3}-\dfrac{3}{20}\) \(x=\dfrac{31}{60}\) Vậy \(x=\dfrac{31}{60}\). e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\) \(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\) \(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\) \(-2x=-\dfrac{73}{20}\) \(x=-\dfrac{73}{20}:\left(-2\right)\) \(x=\dfrac{73}{40}\) Vậy \(x=\dfrac{73}{40}\). K chép lại đề, lm luôn nhé: *\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\) \(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\) \(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\) \(\Rightarrow x=-\dfrac{3}{4}\) * \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\) => K có gt x nào t/m đề * Đề sai * \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\) *\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\) \(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\) \(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\) \(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\) * \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\) \(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\) * \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\) \(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\) \(\Rightarrow-7x=-\dfrac{1}{6}\) \(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\) a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\) \(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\) \(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\) \(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\) b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\) \(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\) \(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\) \(\Rightarrow x\in\varnothing\) c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà. d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\) \(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\) \(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\) e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\) \(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\) \(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\) \(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\) g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) \(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\) \(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\) \(th1:x=0\) \(th2:x=\dfrac{-3}{5}\) h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\) \(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\) \(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\) \(-5x+\dfrac{-1}{2}=2x\) \(\dfrac{-1}{2}=2x+5x\) \(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\) \(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\) \(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\) \(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\) \(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\) \(4\dfrac{1}{5}x=\dfrac{-21}{10}\) \(x=\dfrac{-21}{10}:\dfrac{21}{5}\) \(x=\dfrac{-1}{2}\) 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.
K
Khách
TH
Thanh Hoàng Thanh
5 tháng 3 2022
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\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
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