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a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
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
|\(\frac32x\) + \(\frac12\)| = |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 \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
X : \(3\frac{1}{15}\)=\(1\frac{1}{2}\)
X : \(\frac{46}{15}\)=\(\frac{3}{2}\)
X = \(\frac{3}{2}\)X \(\frac{46}{15}\)
X = \(\frac{23}{5}\)
#)Giải :
\(x\div3\frac{1}{5}=1\frac{1}{2}\)
\(\Leftrightarrow x\div\frac{16}{5}=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{24}{5}\)hay \(4\frac{4}{5}\)
\(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Leftrightarrow\left(\frac{2}{3}-\frac{1}{2}\right)x=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{2}\)
a) = 3/3 x ( -24/54 +45/54 ) : 7/12
= 1 x 21/54 x 12/7
= 18/27
( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )
S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)
S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)
S = \(\frac12\) - \(\frac19\)
S = \(\frac{9}{18}-\frac{2}{18}\)
S = \(\frac{7}{18}\)
1) \(\left(x-1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\right)=0\)
mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
2) \(\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-5}{95}-1=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{95}=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\right)=\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
x - 100 = 1
x = 101
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)

