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Câu I: Ta có:
|5 - 3x| + 2/3=1/6
\(\Rightarrow\) |5 - 3x| =1/6- 2/3
\(\Rightarrow\) |5 - 3x| =-1/2
\(\Rightarrow\)5-3x= -1/2 hoặc 5-3x=1/2
\(\Rightarrow\)x=11/6 hoặc x=3/2.
Câu K: Ta có
- 2,5 + |3x + 5| = -1,5
\(\Rightarrow\) |3x + 5| = -1,5-(- 2,5 )
\(\Rightarrow\) |3x + 5| = 1
\(\Rightarrow\)3x + 5 = 1 hoặc 3x + 5 = -1
\(\Rightarrow\)x= -4/3 hoặc x= -2
a) \(\frac{2}{5}+x=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{2}{5}\)
\(x=\frac{15}{20}-\frac{8}{20}\)
\(x=\frac{7}{20}\)
\(\)b)
\(x-\frac{1}{15}=\frac{3}{10}\\ x=\frac{3}{10}+\frac{1}{15}\\ x=\frac{9}{30}+\frac{2}{30}\\ x=\frac{11}{30}\)
c)
\(\frac{9}{8}-x=\frac{5}{12}\\ x=\frac{9}{8}-\frac{5}{12}\\ x=\frac{27}{24}-\frac{10}{24}\\ x=\frac{17}{24}\)
d)
\(\frac{3}{5}+x=\frac{5}{4}+\frac{7}{10}\\ \frac{3}{5}+x=\frac{25}{20}+\frac{14}{20}\\\frac{3}{5}+x=\frac{39}{20}\\ x=\frac{39}{20}-\frac{3}{5}\\ x=\frac{39}{20}-\frac{12}{20}\\ x=\frac{27}{20} \)
e)
\(\frac{9}{8}-x=\frac{3}{20}+\frac{2}{5}\\ \frac{9}{8}-x=\frac{3}{20}+\frac{8}{20}\\ \frac{9}{8}-x=\frac{11}{20}\\ x=\frac{9}{8}-\frac{11}{20}\\ x=\frac{45}{40}-\frac{22}{40}\\ x=\frac{23}{40}\)
g)
\(x+\frac{1}{3}=\frac{5}{6}+1\frac{7}{10}\\ x+\frac{1}{3}=\frac{5}{6}+\frac{17}{10}\\ x+\frac{1}{3}=\frac{25}{30}+\frac{51}{30}\\ x+\frac{1}{3}=\frac{76}{30}=\frac{38}{15}\\ x=\frac{38}{15}-\frac{1}{3}\\ x=\frac{38}{15}-\frac{5}{15}\\ x=\frac{33}{15}=\frac{11}{5}\)
h)
\(3x-\frac{3}{5}=\frac{1}{2}\\ 3x=\frac{1}{2}+\frac{3}{5}\\ 3x=\frac{5}{10}+\frac{6}{10}\\ 3x=\frac{11}{10}\\ x=\frac{11}{10}:3\\ x=\frac{11}{10}\cdot\frac{1}{3}\\ x=\frac{11}{30}\)
i)
\(4x+\frac{5}{12}+\frac{4}{9}=1\frac{13}{18}\\ 4x+\frac{5}{12}+\frac{4}{9}=\frac{31}{18}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{4}{9}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{8}{18}\\ 4x+\frac{5}{12}=\frac{23}{18}\\ 4x=\frac{23}{18}-\frac{5}{12}\\ 4x=\frac{46}{36}-\frac{15}{36}\\ 4x=\frac{31}{36}\\ x=\frac{31}{36}:4\\ x=\frac{31}{36}\cdot\frac{1}{4}\\ x=\frac{31}{144}\)
k)
\(2-\left(3x+\frac{3}{7}\right)=\frac{9}{21}\\ 2-\left(3x+\frac{3}{7}\right)=\frac{3}{7}\\3x+\frac{3}{7}=2-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{14}{7}-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{11}{7}\\ 3x=\frac{11}{7}-\frac{3}{7}\\ 3x=\frac{8}{7}\\ x=\frac{8}{7}:3\\ x=\frac{8}{7}\cdot\frac{1}{3}\\ x=\frac{8}{21} \)
a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)
b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)
c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)
d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)
a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)
c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)
d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)
e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)
\(\Leftrightarrow-8x+20-2=18x-45\)
\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)