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a) \(1\dfrac{1}{5}+\dfrac{5}{9}+\dfrac{4}{5}+\dfrac{4}{9}\)
\(=\left(\dfrac{6}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\\ =2+1\\ =3\)
b) \(2\dfrac{-7}{10}:\left(\dfrac{5}{7}+\dfrac{3}{14}\right)\)
\(=-\dfrac{27}{10}:\dfrac{13}{14}\\ =-\dfrac{27}{10}\cdot\dfrac{14}{13}\\ =-\dfrac{189}{65}\)
c) \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\\ =\dfrac{7}{30}\)
Giải:
\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\) (Sửa đề)
\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{6}\right)\)
\(\Leftrightarrow A=\dfrac{13}{29}.0\)
\(\Leftrightarrow A=0\)
Vậy ...
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
a: \(=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
b: \(=\dfrac{6+6\cdot4+6\cdot49}{15+15\cdot4+15\cdot49}=\dfrac{6}{15}=\dfrac{2}{5}\)
c: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)
a: \(=\dfrac{-3}{7}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+2+\dfrac{3}{7}=2\)
b: \(=-\dfrac{5}{7}:\left(24-\dfrac{166}{7}\right)+\dfrac{37}{3}\)
\(=-\dfrac{5}{7}:\dfrac{2}{7}+\dfrac{37}{3}=\dfrac{-5}{2}+\dfrac{37}{3}=\dfrac{59}{6}\)
c: \(=4-\dfrac{32}{27}\cdot\dfrac{-27}{8}=4+4=8\)
d: \(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\dfrac{11+5}{20}\cdot\dfrac{5}{7}\)
\(=\dfrac{7}{5}-\dfrac{6}{20}\cdot\dfrac{5}{7}=\dfrac{29}{35}\)
1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)
\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)
\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)