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\(1)\frac{1}{2}x-\frac{3}{5}=\frac{-4}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-4}{5}+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{5}:\frac{1}{2}=\frac{-1}{5}\cdot\frac{2}{1}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}\)
\(2)3\frac{1}{5}-2\frac{1}{3}x=-1\frac{3}{5}+1\frac{7}{10}\)
\(\Rightarrow\frac{16}{5}-\frac{7}{3}x=-\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}-\frac{-8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{16}{5}+\frac{8}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{24}{5}+\frac{17}{10}\)
\(\Rightarrow\frac{7}{3}x=\frac{48}{10}+\frac{17}{10}\)
Đến đây tìm được rồi nhé
3,4, áp dụng bài 1,2 rồi làm :v
a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)
⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)
Vậy: x=0
b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)
⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)
Vậy: x=-1
c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)
⇔\(2x=\frac{3\cdot6}{-1}=-18\)
hay x=-9
Vậy: x=-9
d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)
⇔\(\left(x+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: x∈{2;-4}
e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)
⇔\(9-x=\frac{-12\cdot5}{4}=-15\)
hay x=24
Vậy: x=24
f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)
⇔\(\left(x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy: x∈{5;-3}
g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)
⇔\(\left(5-x\right)^2=4\)
⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy: x∈{3;7}
h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)
⇔\(\left(4-x\right)^2=25\)
⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy: x∈{-1;9}
a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
m) \(\frac{3}{4}-x=\frac{1}{5}\)
\(x=\frac{3}{4}-\frac{1}{5}\)
\(x=\frac{11}{20}\)
Vậy \(x=\frac{11}{20}\)
o) \(\frac{3}{5}x+\frac{1}{4}=\frac{1}{10}\)
\(\frac{3}{5}x=\frac{1}{10}-\frac{1}{4}\)
\(\frac{3}{5}x=-\frac{3}{20}\)
\(x=\left(-\frac{3}{20}\right):\frac{3}{5}\)
\(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}\)
p) \(-\frac{4}{3}x+\frac{3}{2}=\frac{5}{6}\)
\(-\frac{4}{3}x=\frac{5}{6}-\frac{3}{2}\)
\(-\frac{4}{3}x=-\frac{2}{3}\)
\(x=\left(-\frac{2}{3}\right):\left(-\frac{4}{3}\right)\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
m) \(\frac{3}{4}\) - x = \(\frac{1}{5}\)
x= \(\frac{3}{4}\) - \(\frac{1}{5}\)
x= \(\frac{11}{20}\)
o) \(\frac{3}{5}\)x + \(\frac{1}{4}\) = \(\frac{1}{10}\)
\(\frac{3}{5}\)x = \(\frac{1}{10}\) - \(\frac{1}{4}\)
\(\frac{3}{5}\)x = -\(\frac{3}{20}\)
x = \(\frac{-3}{20}\):\(\frac{3}{5}\)
x = \(\frac{-1}{4}\)
p) \(\frac{-4}{3}\)x + \(\frac{3}{2}\) = \(\frac{5}{6}\)
\(\frac{-4}{3}\)x = \(\frac{5}{6}\) - \(\frac{3}{2}\)
\(\frac{-4}{3}\)x = \(\frac{-2}{3}\)
x = \(\frac{-2}{3}\) : \(\frac{-4}{3}\)
x = \(\frac{1}{2}\)
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