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AH
Akai Haruma
Giáo viên
20 tháng 4 2021
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
AH
Akai Haruma
Giáo viên
20 tháng 4 2021
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
A
0
DT
5
28 tháng 1 2016
Ta có\(\frac{-15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)
=>\(\frac{-15}{12}x-\frac{6}{5}x=\frac{1}{2}-\frac{3}{7}\)
=>\(\left(\frac{-15}{12}-\frac{6}{5}\right)x=\frac{1}{14}\)
=>\(\frac{-49}{20}x=\frac{1}{14}\)
=>\(x=\frac{1}{14}:\frac{-49}{20}\)
=>\(x=\frac{-10}{343}\)
13 tháng 2 2020
\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)
=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)
=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)
=>24<28x<231
=>28x\(\in\){25;26;27;28;.............................;230}
=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224
=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}
Vậy x\(\in\) {1;2;3;4;5;6;7;8}
\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)
\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)
\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)
=>3\(⋮\)\(\frac{1}{6}+x\)
=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}
Ta có bảng:
| \(\frac{1}{6}+x\) | -1 | 1 | -3 | 3 |
| x | \(-1\frac{1}{6}\) | \(1\frac{1}{6}\) | \(-3\frac{1}{6}\) | 3\(\frac{1}{6}\) |
Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}
Chúc bn học tốt
NT
8
6 tháng 6 2016
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) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\)
\(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\)
\(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\)
\(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\)
\(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\)
\(\Rightarrow x=-\frac{2}{11}\)
6 tháng 6 2016
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
17 tháng 4 2019
\(a,\left(x\cdot6\frac{2}{7}+\frac{3}{7}\right)\cdot2\frac{1}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(x\cdot\frac{44}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\left(\frac{44x}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{44x+3}{7}\cdot\frac{11}{5}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{11\cdot\left(44x+3\right)}{5\cdot7}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{3}{7}=-2\)
\(\Rightarrow\frac{484x+33}{35}-\frac{15}{35}=-2\)
\(\Rightarrow\frac{484x+33-15}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=-2\)
\(\Rightarrow\frac{484x+18}{35}=\frac{-70}{35}\)
\(\Rightarrow484x+18=\left(-70\right)\)
\(\Rightarrow484x=\left(-70\right)-18\)
\(\Rightarrow484x=-88\)
\(\Rightarrow x=-\frac{88}{484}=-\frac{2}{11}\)
A
17 tháng 4 2019
\(b,x\cdot3\frac{1}{4}+\left(-\frac{7}{6}\right)\cdot x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x\cdot\frac{13}{4}+\left(-\frac{7}{6}\right)\cdot x-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x}{4}+\left(-\frac{7x}{6}\right)-\frac{5}{3}=\frac{5}{12}\)
\(\Rightarrow\frac{13x\cdot3}{12}+\left(-\frac{7x\cdot2}{12}\right)-\frac{5\cdot4}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x}{12}+\left(-\frac{14x}{12}\right)-\frac{20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{39x-14x-20}{12}=\frac{5}{12}\)
\(\Rightarrow\frac{25x-20}{12}=\frac{5}{12}\)
\(\Rightarrow25x-20=5\)
\(\Rightarrow25x=20+5\)
\(\Rightarrow25x=25\)
\(\Rightarrow x=1\)
\(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{2}{3}\right)x=\frac{7}{12}\)
\(\Rightarrow\left(\frac{3}{6}-\frac{4}{6}\right)x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\frac{-1}{6}\)
\(\Rightarrow x=\frac{7}{12}.\left(-6\right)=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
\(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Leftrightarrow\frac{7}{12}x=\frac{7}{12}\)
\(\Leftrightarrow x=1\)