A=\(\dfrac{3^2}{5.14}+\dfrac{3^2}{7.18}+\dfrac{3^2}{9.22}+\dfrac{3^2}{11.26}+\dfrac{3^2}{13.30}\)
Không được sử dụng máy tính!!Trình bày ra dùm mình nha
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3 tháng 6 2019
\(\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{9.22}+\frac{3^2}{11.26}+\frac{3^2}{13.30}\)
\(=3^2.2.\left(\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\right)\)
\(=9.2.\frac{1}{4}.\left(\frac{14-10}{14.10}+\frac{18-14}{14.18}+\frac{22-18}{18.22}+\frac{26-22}{22.26}+\frac{30-26}{26.30}\right)\)
\(=\frac{9}{2}\left(\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
=\(\frac{9}{2}.\left(\frac{1}{10}-\frac{1}{30}\right)=\frac{9}{2}.\frac{1}{15}=\frac{3}{10}\)
XO
3 tháng 6 2019
\(\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{9.22}+\frac{3^2}{13.30}\)
= \(2.\left(\frac{3^2.}{5.2.14}+\frac{3^2}{2.7.18}+\frac{3^2}{2.9.22}+\frac{3^2}{2.13.30}\right)\)
= \(2.\left(\frac{3^2}{10.14}+\frac{3^2}{14.18}+\frac{3^2}{18.22}+\frac{3^2}{26.30}\right)\)
= \(2.\frac{3^2}{4}\left(\frac{4}{10.14}+\frac{4}{14.18}+\frac{4}{18.22}+\frac{4}{26.30}\right)\)
= \(\frac{9}{2}\left(\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{195}\right)\)
= \(\frac{9}{2}.\left(\frac{1}{10}-\frac{1}{22}+\frac{1}{195}\right)\)
= \(\frac{9}{2}.\left(\frac{3}{55}+\frac{1}{195}\right)\)
=\(\frac{9}{2}.\frac{128}{2145}\)
= \(\frac{192}{715}\)
HN
tính: A = \(\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{9.22}+\frac{3^2}{11.26}+\frac{3^2}{13.30}\)
0
HN
1
AK
12 tháng 4 2018
\(\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{99.22}+\frac{3^2}{11.26}+\frac{3^2}{13.30}\)
\(=\frac{9}{5.14}+\frac{9}{7.18}+\frac{9}{9.22}+\frac{9}{11.26}+\frac{9}{13.30}\)
\(=\frac{9}{2}.\left(\frac{4}{10.14}+\frac{4}{14.18}+\frac{4}{18.22}+\frac{4}{22.26}+\frac{4}{26.30}\right)\)
\(=\frac{9}{2}.\left(\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{9}{2}.\left(\frac{1}{10}-\frac{1}{30}\right)\)
\(=\frac{9}{2}.\left(\frac{3}{30}-\frac{1}{30}\right)\)
\(=\frac{9}{2}.\frac{2}{30}\)
\(=\frac{9}{30}\)
\(=\frac{3}{10}\)
Chúc bạn học tốt !!!
4 tháng 5 2017
A =\(\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{9.22}+\frac{3^2}{11.26}+\frac{3^2}{13.30}\)
A =\(3^2.\left(\frac{1}{5.14}+\frac{1}{7.18}+\frac{1}{9.22}+\frac{1}{11.26}+\frac{1}{13.30}\right)\)
A =\(9.\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
A =\(\frac{9}{4}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
A =\(\frac{9}{4}.\left(\frac{1}{5}-\frac{1}{15}\right)\)
A =\(\frac{9}{4}.\frac{2}{15}\)
A =\(\frac{3}{10}\)
AH
Akai Haruma
Giáo viên
26 tháng 10 2023
Lời giải:
$A=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{100}-\sqrt{99}}{(\sqrt{99}+\sqrt{100})(\sqrt{100}-\sqrt{99})}$
$=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+....+\frac{\sqrt{100}-\sqrt{99}}{1}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{100}-\sqrt{99}$
$=\sqrt{100}-1=10-1=9$
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
\(\left\{{}\begin{matrix}x\in\left(0;\dfrac{\pi}{2}\right)\\sinx=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow x=\dfrac{\pi}{3}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{\pi}{6}\Rightarrow cos\dfrac{x}{2}=cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\)
\(A=\dfrac{3^2}{5\cdot14}+\dfrac{3^2}{7\cdot18}+\dfrac{3^2}{9\cdot22}+\dfrac{3^2}{11\cdot26}+\dfrac{3^2}{13\cdot30}\\ =3^2\cdot\left(\dfrac{1}{5\cdot14}+\dfrac{1}{7\cdot18}+\dfrac{1}{9\cdot22}+\dfrac{1}{11\cdot26}+\dfrac{1}{13\cdot30}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}\right)\\ =\dfrac{9}{2}\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\right)\\ =\dfrac{9}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\right)\\ =\dfrac{9}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{15}\right)\\ =\dfrac{9}{4}\cdot\dfrac{2}{15}\\ =\dfrac{3}{10}\)