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a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
(5\(\dfrac{8}{17}\)+\(\dfrac{-4}{17}\)): x +\(\dfrac{22}{7}\):\(\dfrac{52}{3}\)=\(\dfrac{4}{11}\)
\(^{\dfrac{89}{17}}\):x+\(\dfrac{22}{7}\)x\(\dfrac{3}{52}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x+\(\dfrac{33}{182}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x=\(\dfrac{385}{2002}\)
x=\(\dfrac{2314}{85}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
a) 5x + 17 = 2
5x = 2-17
5x =-15
x=-15 : 5
x=-3
b) x+\(\dfrac{7}{8}\)=\(\dfrac{3}{4}\)
x = \(\dfrac{3}{4}-\dfrac{7}{8}\)
x=\(\dfrac{6}{8}-\dfrac{7}{8}\)
x =\(\dfrac{-1}{8}\)