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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: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

2 tháng 8 2020

thank you nha!thanghoa

22 tháng 11 2021

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)

15 tháng 8 2019

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

19 tháng 3 2019

Bài 2:

a) \(\frac{4}{9}+x=\frac{-5}{3}\)

\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)

\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)

Vậy: \(x=\frac{-19}{9}\)

b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)

\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)

\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)

c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)

\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)

28 tháng 2 2020

                                                    Bài giải

a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)

\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)

b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)