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a) (x + 1/2) . (2/3 − 2x) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\)
\(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\)
\(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\)
\(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\)
\(\Rightarrow x=1\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
\(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\)
\(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\)
\(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\)
\(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\)
\(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\)
\(\Rightarrow x\approx28,7\) (số hơi lẻ)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
\(\frac{11}{8}+\frac{13}{6}=\frac{85}{x}\)
\(\frac{85}{24}=\frac{85}{x}\)
\(x=24\)
\(2x-\frac{2}{11}=1\frac{1}{5}\)
\(2x-\frac{2}{11}=\frac{6}{5}\)
\(2x=\frac{76}{55}\)
\(x=\frac{38}{55}\)
a. \(\frac{11}{8}\)+ \(\frac{13}{6}\)= \(\frac{85}{x}\)
\(\frac{85}{24}\) =\(\frac{85}{x}\)
x = \(\frac{24\cdot85}{85}\)
x = 24
b. 2x - \(\frac{2}{11}\)=1\(\frac{1}{5}\)
2x- \(\frac{2}{11}\)= \(\frac{6}{5}\)
2x = \(\frac{6}{5}\)+\(\frac{2}{11}\)
x = \(\frac{76}{55}\): 2
x = \(\frac{38}{55}\)
Học tốt nha ~~~~~ ỌvỌ
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}\)
\(x=-\frac{1}{5}:5=-\frac{1}{25}\)
c) \(\frac{5}{3}-x=7+\frac{4}{5}\)
\(\frac{5}{3}-x=\frac{39}{5}\)
\(x=\frac{5}{3}-\frac{39}{5}=-\frac{92}{15}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}+\frac{5}{11}\)
\(2x=\frac{17}{22}\)
\(x=\frac{17}{22}:2\)
\(x=\frac{17}{44}\)
\(x=-\frac{1}{5}:5\)
NÈ BẠN!!!
a) \(x-\frac{4}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{4}{5}=\frac{25}{35}+\frac{28}{35}=\frac{53}{35}\)
b) \(5x=-\frac{1}{5}+\frac{11}{5}\)
\(5x=2\)
\(x=\frac{2}{5}\)
c)\(\frac{5}{3}-x=7\)
\(x=\frac{5}{3}-7=\frac{5}{3}-\frac{21}{3}=-\frac{16}{3}\)
d) \(-\frac{5}{11}+2x=\frac{7}{22}\)
\(2x=\frac{7}{22}-\frac{-5}{11}=\frac{7}{22}-\frac{-10}{22}=\frac{17}{22}\)
\(x=\frac{17}{22}:2=\frac{17}{22}\cdot\frac{1}{2}=\frac{17}{44}\)
K CHO MÌNH NHA!!!
a, \(\frac{3}{7}=-\frac{x}{21}\Leftrightarrow63=-7x\Leftrightarrow x=-9\)
b, \(\frac{x-1}{6}=\frac{2}{3}\Leftrightarrow\frac{x-1}{6}=\frac{4}{6}\Leftrightarrow x-1=4\Leftrightarrow x=5\)
c, \(\frac{-15}{2x+3}=\frac{-5}{8}\Leftrightarrow-120=-10x-15\Leftrightarrow-10x=-105\Leftrightarrow x=10,5\)( chưa xem lại ko chắc , thử lại giúp mk nha )
a) \(\frac{3}{7}=\frac{-x}{21}\)
\(\Rightarrow-7x=21.3\)
\(-7x=63\)
\(x=63:\left(-7\right)\)
\(x=-9\)
Vậy x=-9
b) \(\frac{x-1}{6}=\frac{2}{3}\)
\(\Rightarrow\left(x-1\right).3=6.2\)
\(3x-3=12\)
\(3x=12+3\)
\(3x=15\)
\(x=15:3\)
\(x=5\)
Vậy x=5
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)