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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.
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) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
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g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
e)\(\frac{1}{3}\left(2x-5\right)=-\frac{2}{3}-\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=-\frac{13}{6}\)
\(2x-5=-\frac{13}{2}\)
\(2x=\frac{-3}{2}\)
\(x=-\frac{3}{4}\)
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!
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a, \(\frac{1}{2}-\frac{3}{5}x=4-\frac{1}{3}x\)
<=> \(\frac{1}{2}-\frac{3}{5}x+\frac{1}{3}x=4\)
<=>\(\frac{1}{2}-x.\left(\frac{3}{5}-\frac{1}{3}\right)=4\)
<=>\(\frac{1}{2}-\frac{4}{15}x=4\)
<=>\(\frac{4}{15}x=\frac{1}{2}-4\)
<=>\(\frac{4}{15}x=\frac{-7}{2}\)
<=> x = \(\frac{-7}{2}:\frac{4}{15}\)
<=> x = \(\frac{-7}{2}.\frac{15}{4}\)
<=> x = \(\frac{-105}{8}\)
b,\(\left(x^2-5\right).x^2=0\)
<=> \(x^2-5=0:x^2\)
<=>\(x^2-5=0\)
<=> \(x^2=5\)
<=> x = 5:x
c, 2 . I x - \(\frac{1}{2}\)I = \(\frac{-1}{3}+5\frac{1}{3}\)
<=>2 . I x - \(\frac{1}{2}\)I = \(\frac{-1}{3}+\frac{5}{3}\)
<=>2 . I x - \(\frac{1}{2}\)I = \(\frac{4}{3}\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{4}{3}:2\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{4}{3}.\frac{1}{2}\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{2}{3}\)
=> x - \(\frac{1}{2}\)= \(\frac{2}{3}\)hoặc x - \(\frac{1}{2}\)= \(\frac{-2}{3}\)
TH1: x -\(\frac{1}{2}\) = \(\frac{2}{3}\)
<=> x = \(\frac{2}{3}\)+ \(\frac{1}{2}\)
<=> x = \(\frac{7}{6}\)
TH2: x - \(\frac{1}{2}\)= \(\frac{-2}{3}\)
<=> x = \(\frac{-2}{3}\)+ \(\frac{1}{2}\)
<=> x = \(\frac{-1}{6}\)
d) I 2x - 3 I - x = 6
=> 2x - 3 - x = 6 hoặc 2x - 3 - x = - 6
TH1:2x - 3 - x = 6
<=> x - 3 = 6
<=> x = 6 + 3
<=> x = 9
TH2: 2x - 3 - x = - 6
<=> x - 3 = -6
<=> x = - 6 + 3
<=> x = - 3
+ I 2x - 3 I