Tính:
B= 1/17 + 7/17x 27 + 7/27x 37 + .....+ 7/1997x 2007
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7-17+27-37+...-1997+2007( có 201 số)
=(7-17)+(27-37)+...+(1987-1997)+2007( có 100 nhóm và 1 số)
=-10+(-10)+...+(-10)+2007( có 100 số -10 và 1 số)
=-10x100+2007
=-1000+2007=1007
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(B=\frac{1}{17}+\frac{7}{17\cdot27}+\frac{7}{27\cdot37}+...+\frac{7}{1997\cdot2007}\)
\(B=\frac{1}{17}+\frac{7}{10}\left(\frac{10}{17\cdot27}+\frac{10}{27\cdot37}+...+\frac{10}{1997\cdot2007}\right)\)
\(B=\frac{1}{17}+\frac{7}{10}\left(\frac{1}{17}-\frac{1}{27}+\frac{1}{27}-\frac{1}{37}+...+\frac{1}{1997}-\frac{1}{2007}\right)\)
\(B=\frac{1}{17}+\frac{7}{10}\left(\frac{1}{17}-\frac{1}{2007}\right)\)
\(B=\frac{1}{17}+\frac{7}{10}\cdot\frac{1990}{34119}\)
\(B=\frac{1}{17}+\frac{1393}{34119}\)
\(B=\frac{200}{2007}\)