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c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)
vậy x = \(\frac{262}{35}\)
d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\)
vậy x = \(\frac{57}{8}\)
e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)
vậy x = \(\frac{1}{4}\)
a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)
vậy x = \(\frac{35}{12}\)
b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)
vậy x = \(\frac{19}{14}\)
Giải:
a) \(\frac{1}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\Leftrightarrow2x=\frac{1}{2}-\left(\frac{1}{5}-\frac{2}{3}\right)\)
\(\Leftrightarrow2x=\frac{1}{2}-\frac{-7}{15}\)
\(\Leftrightarrow2x=\frac{11}{15}\)
\(\Leftrightarrow x=\frac{11}{15}:2\)
\(\Leftrightarrow x=\frac{11}{30}\)
b) \(4\left(\frac{1}{3}-3\right)+\frac{1}{2}=\frac{5}{6}+x\)
\(\Leftrightarrow\frac{-61}{6}=\frac{5}{6}+x\)
\(\Leftrightarrow x=\frac{-61}{6}-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{-66}{6}=-11\)
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) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
a) \(x=\frac{7}{20}\)
b) \(x=\frac{7}{12}\)
c)\(x=\frac{8}{15}\)
a ) \(\frac{7}{8}:x=3-\frac{1}{2}\)
\(\frac{7}{8}:x=\frac{5}{2}\)
\(x=\frac{7}{8}:\frac{5}{2}\)
\(x=0,35\)
b ) \(x+\frac{1}{2}.\frac{1}{3}=\frac{3}{4}\)
\(x+\frac{1}{6}=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{1}{6}\)
\(x=\frac{7}{12}\)
c ) \(\frac{3}{2}.\frac{4}{5}-x=\frac{2}{3}\)
\(\frac{7}{10}-x=\frac{2}{3}\)
\(x=\frac{7}{10}-\frac{2}{3}\)
\(x=\frac{1}{30}\)
d ) \(x.3\frac{1}{3}=3\frac{1}{3}:4\frac{1}{4}\)
\(x:\frac{10}{3}=\frac{10}{3}:\frac{17}{4}\)
\(x:\frac{10}{3}=\frac{40}{51}\)
\(x=\frac{40}{51}:\frac{10}{3}\)
\(x=\frac{4}{17}\)
e ) \(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\)
\(\frac{17}{3}:x=\frac{11}{3}-\frac{5}{2}\)
\(\frac{17}{3}:x=\frac{7}{6}\)
\(x=\frac{17}{3}:\frac{7}{6}\)
\(x=\frac{34}{7}\)
Nếu mình đúng thì các bạn k mình nhé
a) \(\frac{a}{b}x-\frac{7}{8}=\frac{1}{4}\)
\(\Rightarrow\frac{a}{b}x=\frac{1}{4}+\frac{7}{8}\)
\(\Rightarrow\frac{a}{b}x=\frac{9}{8}\)
\(\Rightarrow x=\frac{9}{8}:\frac{a}{b}=\frac{9}{8}.\frac{b}{a}\)
\(\Rightarrow x=\frac{9b}{8a}\)
b) \(\frac{3}{2}x-\frac{1}{2}=\frac{1}{3}:\left(\frac{-5}{6}\right)\)
\(\Rightarrow\frac{3}{2}x-\frac{1}{2}=\frac{-2}{5}\)
\(\Rightarrow\frac{3}{2}x=\frac{-2}{5}+\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\frac{3}{2}\)
\(\Rightarrow x=\frac{1}{15}\)
c) \(\frac{2}{3}\left(x+\frac{5}{4}\right)-\frac{1}{3}\left(\frac{2}{3}-x\right)=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{5}{6}-\frac{2}{9}+\frac{1}{3}x=\frac{4}{3}\)
\(\Rightarrow\frac{2}{3}x+\frac{1}{3}x=\frac{4}{3}-\frac{5}{6}+\frac{2}{9}\)
\(\Rightarrow x=\frac{13}{18}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)
hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)
Vậy: \(x=\frac{9}{10}\)
b) Ta có: \(5\frac{4}{7}:x=13\)
\(\Leftrightarrow\frac{39}{7}:x=13\)
\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)
Vậy: \(x=\frac{3}{7}\)
c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=84\)
\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)
Vậy: x=30
d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)
hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)
Vậy: x=-5
e) Ta có: \(8\frac{2}{3}:x-10=-8\)
\(\Leftrightarrow\frac{26}{3}:x=2\)
hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)
Vậy: \(x=\frac{13}{3}\)
g) Ta có: \(x+30\%=-1.3\)
\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)
hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)
Vậy: \(x=\frac{-8}{5}\)
i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)
\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)
\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)
\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)
Vậy: x=-9
k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)
hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)
Vậy: x=30
m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
CẬU BẤM MÁY TÍNH LÀ RA HẾT
a)
\(x+\frac{11}{4}=\frac{17}{3}\)
\(x=\frac{17}{3}-\frac{11}{4}\)
\(x=\frac{35}{12}\)
b)
\(x-\frac{9}{5}=\frac{23}{7}\)
\(x=\frac{23}{7}+\frac{9}{5}\)
\(x=\frac{178}{35}\)
c)
\(x\)x \(\frac{7}{2}=\frac{19}{4}\)
\(x=\frac{19}{4}:\frac{7}{2}\)
\(x=\frac{19}{14}\)
d)
\(x:\frac{8}{3}=\frac{13}{3}\)
\(x=\frac{13}{3}\)x \(\frac{8}{3}\)
\(x=\frac{104}{9}\)