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\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)
TH1 : \(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=\frac{-1}{2}\)
TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)
\(2x=0+\frac{2}{3}=\frac{2}{3}\)
\(x=\frac{2}{3}\div2=\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{2};\frac{1}{3}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}^2-\frac{1}{5}^2\)
\(x=\frac{2}{5}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)
\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)
\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)
\(=\frac{5}{7}\times\frac{-5}{12}\)
\(=\frac{5\times\left(-5\right)}{7\times12}\)
\(=\frac{-25}{84}\)
\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)
\(=\frac{5}{7}.\frac{-5}{12}\)
\(=-\frac{25}{84}\)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)