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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) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
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 \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
mik ko chép lại đề, mik làm luôn:
a) x - \(\frac{31}{36}=\frac{-13}{38}\)
x = \(\frac{-13}{18}+\frac{31}{36}\)
\(x=\frac{5}{36}\)
b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)
\(\frac{11}{7}-x=\frac{3}{7}\)
x = \(\frac{11}{7}-\frac{3}{7}\)
x = 8/7
c) x + 3/11 = 23/44
x = 23/44 - 3/11
x = 1/4
d) \(\frac{1}{12}-x=\frac{-11}{9}\)
x = \(\frac{1}{12}+\frac{11}{9}\)
x = 47/36
e) \(x-\frac{2}{3}=\frac{-17}{3}\)
x= -17/3 + 2/3
x = -5
f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)
x - 1/2 = 3/4
x = 3/4 + 1/2
x = 5/4
g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)
2x + 3/8 = -3 / 8
2x = -3/8 - 3/8
2x = -9/8
x = -9/8.1/2
x = -9/16
h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)
x - \(\frac{x}{3}=\frac{1}{12}\)
x = \(\frac{1}{12}+\frac{x}{3}\)
x = \(\frac{1+4x}{12}\)
=> 12x = 1+4x
12x - 4x = 1
8x = 1
x = 1/8
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1
(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)
(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)
\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)
(44\(x\) + 3) : \(\frac{11}{5}\) = - 4
44\(x\) + 3 = - 4 x \(\frac{11}{5}\)
44\(x\) + 3 = - \(\frac{44}{5}\)
44\(x\) = - \(\frac{44}{5}\) - 3
44\(x\) = - \(\frac{59}{5}\)
\(x=-\frac{59}{5}:44\)
\(x\) = - \(\frac{59}{250}\)
Vậy \(x\) = - \(\frac{59}{250}\)
A = (19\(\frac58\) : \(\frac{7}{12}\) - 13\(\frac14\): \(\frac72\)) x \(\frac45\) (xem lại đề bài)
B = (9\(\frac{30303}{80808}\) + 7\(\frac{303030}{808080}\)) + 4,03
B = (9\(\frac38\) + 7\(\frac38\)) + 4,03
B =(9 + 7) + (\(\frac38+\frac38\)) + 4,03
B = 16 + 3/4 + 4,03
B = 16 + 0,75 + 4,03
B = 16,75 + 4,03
B = 20,78
a)\(\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}\)
\(=\frac{-7}{25}.1-\frac{18}{25}\)
\(=-\frac{7}{25}-\frac{18}{25}\)
\(=-1\)
b)\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}\)
\(=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0\)
\(=0\)
A . \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}=\frac{-3}{20}\)
B . \(2.x.\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow x.\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow x=0\)
C .\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-7}{20}\)
\(x=\frac{1}{4}:\frac{-7}{20}\)
\(x=\frac{-5}{7}\)
0,7.2\(\frac23\).20.0,375.\(\frac{5}{28}\)
= \(\frac{7}{10}\) .\(\frac83\).20.\(\frac38\).\(\frac{5}{28}\)
= (\(\frac{7}{10}\).\(\frac{5}{28}\).20).(\(\frac83\).\(\frac38\))
= (\(\frac18.20\)). 1
= \(\frac52\)
B = (9,75 .21\(\frac37\) + \(\frac{39}{4}\).18\(\frac47\))
B = (\(\frac{39}{4}\).\(\frac{150}{7}\frac{}{7}\)+ \(\frac{39}{4}\).\(\frac{130}{7}\))
B = \(\frac{39}{4}\).(\(\frac{150}{7}\) + \(\frac{130}{7}\))
B = \(\frac{39}{4}\).\(\frac{280}{7}\)
B = 390
Bài 2 tìm x:
(x + \(\frac12\)).(\(\frac23\) - 2x) = 0
x + \(\frac12\) = 0 hoặc \(\frac23\) - 2x = 0
x + \(\frac12\) = 0
x = - \(\frac12\)
\(\frac23\) - 2x = 0
2x = \(\frac23\)
x = \(\frac23\) : 2
x = \(\frac13\)
Vậy x ∈{- \(\frac12\); \(\frac13\)}
\(\frac23\)x - \(\frac12\)x = \(\frac{5}{12}\)
x.(\(\frac23\) - \(\frac12\)) = \(\frac{5}{12}\)
x.\(\frac16\) = \(\frac{5}{12}\)
x = \(\frac{5}{12}\) : \(\frac16\)
x = \(\frac{5}{12}\) x 6
x = \(\frac52\)
Vậy x = \(\frac52\)
x.3\(\frac14\) + (-\(\frac76\)).x - 1\(\frac23\) = \(\frac{5}{12}\)
x.\(\frac{13}{4}\) - \(\frac76\)x - \(\frac53\) = \(\frac{5}{12}\)
x.(\(\frac{13}{4}\) - \(\frac76\)) = \(\frac{5}{12}\) + \(\frac53\)
x.(\(\frac{39}{12}\) - \(\frac{14}{12}\)) = \(\frac{5}{12}\) + \(\frac{20}{12}\)
x.\(\frac{25}{12}\) = \(\frac{25}{12}\)
x = \(\frac{25}{12}\) : \(\frac{25}{12}\)
x = \(\frac{25}{12}\) x \(\frac{12}{25}\)
x = \(1\)
Vậy x = 1
5\(\frac{8}{17}\) : \(x\) + (-\(\frac{4}{17}\)) : \(x\) + 3\(\frac17\) : 17\(\frac13\) = \(\frac{4}{11}\)
\(\frac{93}{17}\).\(\frac{1}{x}\) - \(\frac{4}{17}\) .\(\frac{1}{x}\) + \(\frac{22}{7}\) : \(\frac{52}{3}\) = \(\frac{4}{11}\)
\(\frac{1}{x}\).(\(\frac{93}{17}\) - \(\frac{4}{17}\)) + \(\frac{22}{7}\).\(\frac{3}{52}\) = \(\frac{4}{11}\)
\(\frac{1}{x}\).\(\frac{89}{17}\) + \(\frac{33}{182}\) = \(\frac{4}{11}\)
\(\frac{1}{x}\).\(\frac{89}{17}\) = \(\frac{4}{11}\) - \(\frac{33}{182}\)
\(\frac{1}{x}\).\(\frac{89}{17}\) = \(\frac{728}{2002}\) - \(\frac{363}{2002}\)
\(\frac{1}{x}\frac{.89}{17}\) = \(\frac{365}{2002}\)
\(x\) = \(\frac{89}{17}\) : \(\frac{365}{2002}\)
\(x\) = \(\frac{89}{17}\) x \(\frac{2002}{365}\)
\(x\) = \(\frac{178178}{6205}\)
Vậy \(x\) = \(\frac{178178}{6205}\)