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a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)
A = (19\(\frac58\) : \(\frac{7}{12}\) - 13\(\frac14\): \(\frac72\)) x \(\frac45\) (xem lại đề bài)
B = (9\(\frac{30303}{80808}\) + 7\(\frac{303030}{808080}\)) + 4,03
B = (9\(\frac38\) + 7\(\frac38\)) + 4,03
B =(9 + 7) + (\(\frac38+\frac38\)) + 4,03
B = 16 + 3/4 + 4,03
B = 16 + 0,75 + 4,03
B = 16,75 + 4,03
B = 20,78
\(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
\(\Leftrightarrow5\left(7-8x\right)=6\left(-4+2x\right)\)
\(\Leftrightarrow35-40x=-24+12x\)
\(\Leftrightarrow35+24=40x+12x\)
\(\Leftrightarrow59=52x\)
\(\Rightarrow x=\frac{59}{52}\)
\(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
\(\Leftrightarrow1-3:x=8\)
\(\Leftrightarrow-3:x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
\(\frac{x}{5}=\frac23\)
\(x\) = \(\frac23\times5\)
\(x=\frac{10}{3}\)
Vậy \(x=\frac{10}{3}\)
\(\frac{x}{3}-\frac12=\frac15\)
\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)
\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(x=\frac{7}{10}\times3\)
\(x=\frac{21}{10}\)
Vậy \(x=\frac{21}{10}\)
\(\frac{x}{5}+\frac12=\frac{6}{10}\)
\(\frac{x}{5}=\frac{6}{10}-\frac12\)
\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)
\(\frac{x}{5}=\frac{1}{10}\)
\(x=\frac{1}{10}\times5\)
\(x=\frac12\)
Vậy \(x=\frac12\)
\(\frac{x+3}{15}\) = \(\frac13\)
\(x+3=\frac13\times15\)
\(x+3=5\)
\(x=5-3\)
\(x=2\)
Vậy \(x=2\)
|\(\frac32x\) + \(\frac12\)| = |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
thiếu đề
vế phải = bao nhiu
Hả sao ko có
phần 1 dấu + chuyển thành dâu =
phai bang so nao chu
hay phai co dau :<,>,= thi minh lam duoc