1+2+3+4+5+6+...=- 1/12 Mọi người có tin ko?
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a,\(\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}\)
=\(\frac{1}{2}+\frac{3}{4}+\frac{-3}{4}+\frac{4}{5}\)
=\(\left(\frac{3}{4}+\frac{-3}{4}\right)+\frac{1}{2}+\frac{4}{5}\)
= \(0+\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
Nhiều quá bạn ơi
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
a. \(1+\frac{2x-5}{6}=\frac{3-x}{4}\)
\(\Leftrightarrow\frac{12}{12}+\frac{2\left(2x-5\right)}{12}-\frac{3\left(3-x\right)}{12}=0\)
\(\Leftrightarrow12+4x-10-9+3x=0\)
\(\Leftrightarrow7x-7=0\)
\(\Leftrightarrow7x=7\Leftrightarrow x=1\)
b. \(\frac{x+1}{2}-\frac{x-2}{3}=\frac{3\left(x+1\right)}{6}-\frac{2\left(x-2\right)}{6}=\frac{x+7}{6}\)
c. \(\frac{2x-1}{3}+x=\frac{x+4}{2}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}+\frac{6x}{6}-\frac{3\left(x+4\right)}{6}=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
d. \(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}+\frac{2x-1}{12}\)
\(=\frac{6\left(x+5\right)}{24}-\frac{8\left(2x-3\right)}{24}-\frac{3\left(6x-1\right)}{24}+\frac{2\left(2x-1\right)}{24}\)
\(=6x+30-16x+24-18x+3+4x-2\)
\(=-24x-55\)
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