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\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)

a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)

\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)

Câu 1:
a: \(\left(3x-15\right)=3^7:3^5\)
=>3x-15=9
=>3x=24
hay x=8
b: \(\left(4x+32\right)=43\cdot2^2\)
=>4x+32=172
=>4x=140
hay x=35
c: \(6^{2x-7}=216\)
=>2x-7=3
=>2x=10
hay x=5
d: \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x\cdot26=650\)
\(\Leftrightarrow5^x=25\)
hay x=2

\((2,7.x-1\frac{1}{2})\div\frac{2}{7}=\frac{-21}{4}\) \(3\frac{1}{3}.x+16\frac{3}{4}=-13.25\)
\(2,7.x-1\frac{1}{2}=-\frac{21}{4}\cdot\frac{2}{7}\) \(\frac{10}{3}.x+\frac{67}{4}=-13.25\)
\(2,7.x-\frac{3}{2}=-\frac{3}{2}\) \(\frac{10}{3}.x+\frac{67}{4}=-\frac{53}{4}\)
\(2,7.x=-\frac{3}{2}+\frac{3}{2}\) \(\frac{10}{3}.x=-\frac{53}{4}-\frac{67}{4}\)
\(2,7.x=0\) \(\frac{10}{3}.x=-30\)
\(x=0:2,7\) \(x=-30:\frac{10}{3}\)
\(x=0\) \(x=-9\)
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
\(a,\left(4+6\times x\right)\times5=290\)
\(4+6\times x=290:5\)
\(4+6\times x=58\)
\(6\times x=54\)
\(x=9\)
\(b,428\times x-12\times x-18\times x=5208000\)
\(\left(428-12-18\right)\times x=5208000\)
\(398\times x=5208000\)
\(x=\dfrac{5208000}{398}\)
\(c,5\times x+3,75\times x+1,25\times x=20\)
\(x\times\left(5+3,75+1,25\right)=20\)
\(x\times10=20\)
\(x=2\)
\(d,280:\left(7+3\times2\right)=4\)
Đề thiếu bạn nhé.
\(f,6\times\left(28-8\times x\right)=72\)
\(28-8\times x=12\)
\(8\times x=16\)
\(x=2\)
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