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a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)
b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)
Xảy ra 2 trường hợp:
+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)
+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)
Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)
a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)
Vậy \(x=-6\)
b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)
Chúc bạn học tốt!!!
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: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
\(a)\dfrac{4}{7}+\dfrac{3}{7}\div x=-5\)
\(\dfrac{3}{7}\div x=\left(-5\right)-\dfrac{4}{7}\)
\(\dfrac{3}{7}\div x=\dfrac{-39}{7}\)
\(x=\dfrac{3}{7}\div\dfrac{-39}{7}\)
\(x=\dfrac{-1}{13}\)
\(Vậy\) \(x=\dfrac{-1}{13}\)
\(b)\dfrac{-1}{2}\times x+\dfrac{2}{3}\times\dfrac{-5}{4}=\dfrac{3}{5}-\dfrac{5}{6}\)
\(\dfrac{-1}{2}\times x+\dfrac{-5}{6}=\dfrac{-7}{30}\)
\(\dfrac{-1}{2}\times x=\dfrac{-7}{30}-\dfrac{-5}{6}\)
\(\dfrac{-1}{2}\times x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}\div\dfrac{-1}{2}\)
\(x=\dfrac{-6}{5}\)
\(Vậy\) \(x=\dfrac{-6}{5}\)
\(\dfrac{4}{7}+\dfrac{3}{7}:x=-5\\ \dfrac{3}{7}:x=-5-\dfrac{4}{7}\\ \dfrac{3}{7}:x=\dfrac{-39}{7}\\ x=\dfrac{3}{7}:\dfrac{-39}{7}\\ x=\dfrac{-1}{13}\)
a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)
x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)
x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)
x=\(\dfrac{63}{142}\)
a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)
\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)
\(x.\dfrac{71}{42}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{71}{42}\)
\(x=\dfrac{63}{142}\)
b\(\dfrac{5}{4}-\dfrac{3}{5}:x=1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{5}{4}-1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{-1}{12}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{12}\)
\(x=\dfrac{-36}{5}\)
c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)
\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)
\(\dfrac{4}{7}x-\dfrac{1}{3}=\dfrac{1}{7}\)
\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{10}{21}\)
\(x=\dfrac{10}{21}:\dfrac{4}{7}\)
\(x=\dfrac{5}{6}\)
d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)
\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)
\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)
\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)
\(x=\dfrac{27}{110}\)
có sai sót j xin bn thông cảm !
a.\(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)
\(\dfrac{5}{3}x-2\dfrac{1}{3}=\dfrac{5}{6}\)
\(\dfrac{5}{3}x=\dfrac{5}{6}+2\dfrac{1}{3}\)
\(\dfrac{5}{3}x=\dfrac{19}{6}\)
\(x=\dfrac{19}{5}:\dfrac{5}{3}\)
\(x=\dfrac{19}{10}\)
b. \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)
\(2\dfrac{1}{6}:x+\dfrac{5}{8}=\dfrac{47}{63}\)
\(2\dfrac{1}{6}:x=\dfrac{47}{63}-\dfrac{5}{8}\)
\(2\dfrac{1}{6}:x=\dfrac{61}{504}\)
\(x=2\dfrac{1}{6}:\dfrac{61}{504}\)
\(x=\dfrac{1092}{61}\)
a, \(\dfrac{5}{3}x-2\dfrac{1}{3}=-\dfrac{4}{3}.1\dfrac{1}{8}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{3}x=-\dfrac{3}{2}-\dfrac{2}{3}+2\dfrac{1}{3}=-\dfrac{3}{2}+2=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{5}{3}=\dfrac{3}{10}\)
b, \(2\dfrac{1}{6}:x-\dfrac{-5}{8}=\dfrac{-7}{15}:4\dfrac{1}{5}-\dfrac{-6}{7}\)
\(\Rightarrow\dfrac{13}{6}:x=-\dfrac{7}{15}:\dfrac{21}{5}+\dfrac{6}{7}-\dfrac{5}{8}\)
\(\Rightarrow\dfrac{13}{6}:x=\dfrac{61}{504}\Rightarrow x=\dfrac{1092}{61}\)
c, \(\left(\dfrac{5}{6}x-0,3\right):2\dfrac{1}{3}=25\%\)
\(\Rightarrow\dfrac{5}{6}x-0,3=\dfrac{1}{4}.\dfrac{7}{3}\)
\(\Rightarrow\dfrac{5}{6}x=\dfrac{7}{12}+0,3=\dfrac{53}{60}\)
\(\Rightarrow x=\dfrac{53}{60}:\dfrac{5}{6}=1,06\)
d, \(\dfrac{4}{7}-\dfrac{2}{3}x=1,5+\dfrac{4}{5}x\)
\(\Rightarrow\dfrac{4}{5}x+\dfrac{2}{3}x=\dfrac{4}{7}-1,5\)
\(\Rightarrow\dfrac{22}{15}x=-\dfrac{13}{14}\Rightarrow x=-\dfrac{195}{308}\)
Chúc bạn học tốt!!!
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
a, \(\dfrac{3}{4}.x+\dfrac{1}{5}=\dfrac{1}{6}\)
\(\dfrac{3}{4}.x=\dfrac{1}{6}-\dfrac{1}{5}\)
\(\dfrac{3}{4}.x=\dfrac{-1}{30}\)
\(x=\dfrac{-1}{30}:\dfrac{3}{4}\)
\(x=\dfrac{-2}{45}\)
b, \(\left(4\dfrac{1}{2}-\dfrac{2}{5}x\right):\dfrac{7}{4}=\dfrac{11}{14}\)
\(\left(4\dfrac{1}{2}-\dfrac{2}{5}x\right)=\dfrac{11}{14}.\dfrac{7}{4}\)
\(4\dfrac{1}{2}-\dfrac{2}{5}x=\dfrac{11}{8}\)
\(\dfrac{2}{5}x=4\dfrac{1}{2}-\dfrac{11}{8}\)
\(\dfrac{2}{5}x=\dfrac{9}{2}-\dfrac{11}{8}\)
\(\dfrac{2}{5}x=\dfrac{25}{8}\)
\(x=\dfrac{25}{8}:\dfrac{2}{5}\)
\(x=\dfrac{125}{16}\)
Lời giải:
a)
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}:\frac{4}{7}=\frac{91}{60}$
b)
$\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}$
$\frac{5}{7}:x=\frac{-19}{30}$
$x=\frac{5}{7}:\frac{-19}{30}=\frac{-150}{133}$
a) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
\(x=\dfrac{91}{60}\)
b) \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\)
\(\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\dfrac{5}{7}:x=\dfrac{-19}{30}\)
\(x=\dfrac{5}{7}:\dfrac{-19}{30}\)
\(x=\dfrac{-150}{133}\)