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\(A.3x-\frac{3}{2}-5x+\frac{10}{3}=1\)
\(3x-\frac{3}{2}-5x=-\frac{7}{3}\)
\(3x-5x=-\frac{7}{3}+\frac{3}{2}\)
\(-2x=-\frac{5}{6}\)
\(x=-\frac{5}{6}:\left(-2\right)\)
\(x=\frac{5}{12}\)
a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)
\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=4\frac{1}{4}\)
\(\frac{1}{10}x=\frac{17}{4}+\frac{3}{8}\)
\(\frac{1}{10}x=\frac{37}{8}\)
\(x=\frac{185}{4}\)
\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)
\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)
\(\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)
\(\frac{1}{10}x=\frac{37}{8}\)
\(x=\frac{37}{8}:\frac{1}{10}\)
\(x=\frac{185}{4}\)
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
sory,de qua dai nen minh lam kha lau