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A = \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + \(\dfrac{2}{7.9}\) + ... + \(\dfrac{2}{19.21}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + ... + \(\dfrac{1}{19}\) - \(\dfrac{1}{21}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{21}\)
A = \(\dfrac{2}{7}\)
\(\dfrac{3}{5}x-\dfrac{2}{3}x-\dfrac{1}{7}x=-\dfrac{5}{21}\\ \Rightarrow-\dfrac{22}{105}x=-\dfrac{5}{21}\\ \Rightarrow x=\dfrac{25}{22}\)
\(a,\frac{8^{12}.5^{21}}{2^{17}.10^{19}}=\frac{\left(2^3\right)^{12}.5^{21}}{2^{17}.2^{19}.5^{19}}=\frac{2^{36}.5^{21}}{2^{36}.5^{19}}=25\)
\(b,\left(x-5\right).\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow x-5=0\)hoặc \(x+\frac{1}{2}=0\)
\(x=5\)hoặc \(x=-\frac{1}{2}\)
\(c,\left|x-6\right|-\frac{1}{2}=\frac{3}{2}\)
\(\left|x-6\right|=2\)
\(\Rightarrow x-6=2\)hoặc \(x-6=-2\)
\(x=8\)hoặc \(x=4\)
\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(\frac{15}{34}+\frac{1}{3}+\frac{19}{34}+\left(-\frac{4}{3}\right)+\frac{3}{7}\)
\(\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{1}{3}-\frac{4}{3}\right)+\frac{3}{7}\)
\(1+\left(-1\right)+\frac{3}{7}\)
\(=0+\frac{3}{7}=\frac{3}{7}\)
\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=\)\(\frac{15}{34}+\frac{1}{3}+\frac{19}{34}+\left(-\frac{4}{3}\right)+\frac{3}{7}\)
\(=\)\(\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{1}{3}-\frac{4}{3}\right)+\frac{3}{7}\)
\(=\)\(1+\left(-1\right)+\frac{3}{7}\)
\(=\)\(0+\frac{3}{7}\)
\(=\)\(\frac{3}{7}\)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
Ta có :
\(\frac{x}{3}=\frac{y}{5}\) (1)
\(\left(x+2\right)+\left(y+3\right)=21\)
\(\Leftrightarrow x+2+y+3=21\)
\(\Leftrightarrow x+y=16\)
\(\Leftrightarrow x=16-y\)
Thay x = 16 - y vào (1) ta được :
\(\frac{16-y}{3}=\frac{y}{5}\)
\(\Rightarrow5\left(16-y\right)=3y\)
\(\Leftrightarrow80-5y=3y\)
\(\Leftrightarrow80=8y\)
\(\Leftrightarrow y=10\)
\(\Rightarrow x=16-10=6\)
Vậy x = 6 và y = 10
\(x.\frac{-3}{7}=\frac{5}{21}\)
\(\Leftrightarrow x=\frac{5}{21}:\frac{-3}{7}\)
\(\Leftrightarrow x=\frac{5}{21}.\frac{7}{-3}\)
\(\Leftrightarrow x=-\frac{5}{9}\)
hok tốt!!
Ta có : \(x.\left(\frac{-3}{7}\right)=\frac{5}{21}\)
\(\Leftrightarrow x=\frac{5}{21}-\frac{-3}{7}\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)