\(1,5+1\frac{1}{4}\times x=\frac{2}{3}\)
Tìm x:\(\left(2,7.x-1\frac{1}{2}.x\right)\div\frac{2}{7}=\frac{-21}{4}\)
\(3\frac{1}{3}\times x+16\frac{3}{4}=-13,25\)
\(\left(4,5-2\times x\right)\div\frac{3}{4}=1\frac{1}{3}\)
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\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)
a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
\((2,7.x-1\frac{1}{2})\div\frac{2}{7}=\frac{-21}{4}\) \(3\frac{1}{3}.x+16\frac{3}{4}=-13.25\)
\(2,7.x-1\frac{1}{2}=-\frac{21}{4}\cdot\frac{2}{7}\) \(\frac{10}{3}.x+\frac{67}{4}=-13.25\)
\(2,7.x-\frac{3}{2}=-\frac{3}{2}\) \(\frac{10}{3}.x+\frac{67}{4}=-\frac{53}{4}\)
\(2,7.x=-\frac{3}{2}+\frac{3}{2}\) \(\frac{10}{3}.x=-\frac{53}{4}-\frac{67}{4}\)
\(2,7.x=0\) \(\frac{10}{3}.x=-30\)
\(x=0:2,7\) \(x=-30:\frac{10}{3}\)
\(x=0\) \(x=-9\)
Vậy x=0 Vậy x= -9
\(\left(4.5-2.x\right):\frac{3}{4}=1\frac{1}{3}\) \(1.5+1\frac{1}{4}.x=\frac{2}{3}\)
\(\left(4.5-2.x\right)=1\frac{1}{3}\cdot\frac{3}{4}\) \(1\frac{1}{4}.x=\frac{2}{3}-1.5\)
\(4.5-2.x=\frac{4}{3}\cdot\frac{3}{4}\) \(\frac{5}{4}.x=\frac{2}{3}-\frac{3}{2}\)
\(4.5-2.x=1\) \(\frac{5}{4}.x=-\frac{5}{6}\)
\(2.x=4.5-1\) \(x=-\frac{5}{6}:\frac{5}{4}\)
\(2.x=3.5\) \(x=-\frac{2}{3}\)
\(x=3.5:2\)
\(x=1.75\) Vậy \(x=-\frac{2}{3}\)
Vậy x=1.75