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1 tháng 8 2020
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: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
18 tháng 7 2016
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 = \(\frac{1}{4}\)
18 tháng 7 2016
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 = \(\frac{19}{14}\)
DK
25 tháng 8 2018
Bài 1
a)\(\left(-\dfrac{2}{3}\right).\dfrac{3}{11}-\left(\dfrac{4}{3}\right)^2.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\left(\dfrac{4}{3}\right)^2\right]\)
\(=\dfrac{3}{11}.\left[\left(-\dfrac{2}{3}\right)-\dfrac{4}{3}.\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left[\left(-2\right).\dfrac{4}{3}\right]\)
\(=\dfrac{3}{11}.\left(-\dfrac{8}{3}\right)\)
\(=-\dfrac{24}{33}\)
5 tháng 8 2017
a) \(x=\frac{7}{20}\)
b) \(x=\frac{7}{12}\)
c)\(x=\frac{8}{15}\)
NT
5 tháng 8 2017
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
MV
23 tháng 4 2017
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
24 tháng 7 2020
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
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 => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
9 tháng 8 2017
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 \(\frac{8}{3}\)
\(x=\frac{104}{9}\)
Z
13 tháng 8 2019
a, Ta có:
\(2.x-\frac{1}{2}=\frac{5}{4}-\frac{1}{2}.x\)
\(\Leftrightarrow2.x+\frac{1}{2}.x=\frac{5}{4}+\frac{1}{2}\)
\(\Leftrightarrow\frac{5}{2}.x=\frac{7}{4}\)
\(\Leftrightarrow x=\frac{7}{10}\). Vậy: \(x=\frac{7}{10}\)
b, Ta có:
\(\frac{3}{4}-\frac{5}{2}.x=2-3.x\)
\(\Leftrightarrow2-\frac{3}{4}=3.x-\frac{5}{2}.x\)
\(\Leftrightarrow\frac{5}{4}=\frac{1}{2}.x\)
\(\Leftrightarrow x=\frac{5}{2}\). Vậy: \(x=\frac{5}{2}\)
c, Ta có:
\(\frac{3}{2}.\left(x-\frac{1}{3}\right)-2.\left(x-\frac{1}{2}\right)=3\)
\(\Leftrightarrow\frac{3}{2}.x-\frac{3}{2}.\frac{1}{3}-\left(2.x-2.\frac{1}{2}\right)=3\)
\(\Leftrightarrow\frac{3}{2}.x-\frac{1}{2}-2.x+1=3\)
\(\Leftrightarrow-\frac{1}{2}.x+\frac{1}{2}=3\)
\(\Leftrightarrow-\frac{1}{2}.x=\frac{5}{2}\)
\(\Leftrightarrow x=-5\). Vậy: \(x=-5\)
d, Ta có:
\(\frac{4}{3}.\left(x-\frac{3}{2}\right)-3.\left(\frac{1}{3}.x+1\right)=6\)
\(\Leftrightarrow\frac{4}{3}.x-\frac{4}{3}.\frac{3}{2}-3.\frac{1}{3}.x-3=6\)
\(\Leftrightarrow\frac{4}{3}.x-2-x-3=6\)
\(\Leftrightarrow\frac{1}{3}.x-5=6\)
\(\Leftrightarrow\frac{1}{3}.x=11\)
\(\Leftrightarrow x=33\). Vậy: \(x=33\)
Chúc bạn học tốt!
Tick cho mình nhé!
9 tháng 8 2017
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)
\(=\frac{1.2.3......2016}{2.3.4.......2017}\)
\(=\frac{1}{2017}\)
\(\left(1+2+3+4\right)^2\cdot\left(1+x\right)^{2021}=1^3+2^3+3^3+4^3\\ \Rightarrow10^2\cdot\left(1+x\right)^{2021}=1+8+27+64\\ \Rightarrow100\cdot\left(1+x\right)^{2021}=100\\ \Rightarrow\left(1+x\right)^{2021}=100:100\\ \Rightarrow\left(1+x\right)^{2021}=1\\ \Rightarrow1+x=1\\ \Rightarrow x=1-1\\ \Rightarrow x=0\)