1+2+3+...+14=?
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\(\frac{1}{2}\cdot\frac{18}{14}+\frac{1}{2}\cdot\frac{3}{14}-\frac{1}{2}\cdot\frac{7}{14}\)
\(=\frac{1}{2}\left(\frac{18}{14}+\frac{3}{14}-\frac{7}{14}\right)\)
\(=\frac{1}{2}\cdot1=\frac{1}{2}\)
\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)
a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)
Đặt A=1+14+142+143+...+1414
A=1+14+142+143+...+1414=(1+14)+(142+143)+...+(1413+1414)
A=(1+14)+142x(1+14)+...+1413x(1+14)
A=15+142x15+...+1413x15
A=15x(1+142+...+1413)
Vì 15 chia hết cho 3=>Vậy A chia hết cho 3(dpcm)
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)
Vậy A<\(\frac{1}{2}\).
b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
Vậy \(B< 1\).
\(=105\)
\(=105\)