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|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
Bài 2:
- \(\frac75\) + \(x\) = - \(\frac34\)
\(x\) = - \(\frac34+\frac75\)
\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)
\(x\) = \(\frac{13}{20}\)
Vậy \(x=\frac{13}{20}\)
Bài 3:
- \(\frac74\) + \(\frac14:x\) = \(\frac32\)
\(\frac14:x\) = \(\frac32+\frac74\)
\(\frac14:x\) = \(\frac64+\frac74\)
\(\frac14:x=\frac{13}{4}\)
\(x=\frac14:\frac{13}{4}\)
\(x=\frac14\times\frac{4}{13}\)
\(x=\frac{1}{13}\)
Vậy \(x=\frac{1}{13}\)
a, \(\frac{x}{5}=\frac{2}{3}\Leftrightarrow x=\frac{10}{3}\)
b, \(\frac{x}{-24}=\frac{20}{42}\Leftrightarrow x=-\frac{80}{7}\)
c, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow3x+9=15\Leftrightarrow x=2\)
d, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\Leftrightarrow x\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\Leftrightarrow-\frac{5}{6}x=\frac{5}{12}\Leftrightarrow x=-\frac{1}{2}\)
a) \(x+\frac{5}{12}=-1\frac{2}{7}\)
\(\Leftrightarrow x+\frac{5}{12}=\frac{-9}{7}\)
\(\Leftrightarrow x=\frac{-143}{84}\)
Vậy ...
b) \(4\frac{1}{2}x:\frac{5}{12}=0,5\)
\(\Leftrightarrow\frac{9}{2}x=\frac{5}{24}\)
\(\Leftrightarrow x=\frac{5}{108}\)
vậy...
c) \(7,5.1\frac{3}{4}x=6\frac{2}{5}\)
\(\Leftrightarrow\frac{105}{8}x=\frac{32}{5}\)
\(\Leftrightarrow x=\frac{256}{525}\)
Vậy ...
Câu 1a:
75% - 1 1/2 + 0,5 : 5/12 - (-1/2)^2
= 0,75 - 1,5 + 0,5 x 12/5 - 0,25
= (0,75 - 0,25) - 1,5 + 1,2
= 0,5 - 1,5 + 1,2
= - 1 + 1,2
= 0,2
Bài 1b:
(5/7.0,6 - 5: 3 1/2). (40% - 1,4).(-2)^3
= (3/7 - 5 x 2/7).(0,4 - 1,4).8
= - 1.(-1).8
= 8