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\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=\frac{891}{100}\)
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)=9.\frac{99}{100}=\frac{891}{100}\)
\(\frac{x-7}{3}=\frac{-27}{7-x}\)
\(\Leftrightarrow\left(x-7\right)\left(7-x\right)=\left(-27\right).3\)
\(\Leftrightarrow7x-x^2-49+7x=-81\)
\(\Leftrightarrow-x^2+14x-49=-81\)
\(\Leftrightarrow-x^2+14x+32=0\)
\(\Leftrightarrow x^2-14x-32=0\)
\(\Leftrightarrow x^2+2x-16x-32=0\)
\(\Leftrightarrow x\left(x+2\right)-16\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=16\end{cases}}}\)
Tổng trên = 1-2^2/2^2 . 1-3^2/3^2 . ..... . 1-100^2/100^2
= -(2^2-1/2^2 . 3^2-1/3^2 . ...... . 100^2-1/100^2 )
= -(1.3/2^2 . 2.4/3^2 . ..... . 99.101/100)
= -(1.2.3. .... .99 . 3.4.5. ... .101 / 2.3.4 . ... . 100 . 2.3.4 . ..... . 100)
= -(1.2.3. ... . 99/2.3.4. .... .100) . (3.4.5. .... .101/2.3.4 . .... . 100)
= -1/100 . 101/2 = -101/200
Tk mk nha
(-\(\frac{4}{12}\)) +7=(\(-\frac{1}{3}\)) + 7
=(-\(\frac{1}{3}\))+\(\frac{21}{3}\)
=\(\frac{20}{3}\)
Đặt A là tên của biểu thức trên
2A = \(\frac{7.2}{5.9}+\frac{7.2}{9.11}+\frac{7.2}{11.13}+\frac{7.2}{13.15}+...+\frac{7.2}{2015.2017}\)
2A = \(7\left(\frac{2}{5.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{2015.2017}\right)\)
2A = \(7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
2A = \(7\left(\frac{1}{5}-\frac{1}{2017}\right)\)
2A = \(7\cdot\frac{2012}{10085}\)
2A = \(\frac{14084}{10085}\)
A = \(\frac{14084}{10085}:2\)
A = \(\frac{7042}{10085}\)
\(\frac{7}{5.9}+\frac{7}{9.11}+\frac{7}{11.13}+\frac{7}{11.13}+...+\frac{7}{2015.2017}\)
\(=\frac{7}{5.9}+\frac{7}{2}.\left(\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{7}{45}+\frac{7}{2}.\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)