b1 tính
A= 3/5x2+3//8x11x3/11x14
K
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T
3
10 tháng 7 2019
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
10 tháng 7 2019
\(A=\)\(\frac{3}{5\times2}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{11\times14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
PH
1
DD
Đoàn Đức Hà
Giáo viên
28 tháng 9 2021
a) \(\frac{3}{4\times9}+\frac{3}{9\times14}+...+\frac{3}{54\times59}+\frac{3}{59\times64}\)
\(=\frac{3}{5}\times\left(\frac{5}{4\times9}+\frac{5}{9\times14}+...+\frac{5}{59\times64}\right)\)
\(=\frac{3}{5}\times\left(\frac{9-4}{4\times9}+\frac{14-9}{9\times14}+...+\frac{64-59}{59\times64}\right)\)
\(=\frac{3}{5}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{59}-\frac{1}{64}\right)\)
\(=\frac{3}{5}\times\left(\frac{1}{4}-\frac{1}{64}\right)\)
\(=\frac{9}{64}\)
b) \(\frac{2}{8\times11}+\frac{2}{11\times14}+...+\frac{2}{23\times26}+\frac{2}{26\times29}\)
\(=\frac{2}{3}\times\left(\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{26\times29}\right)\)
\(=\frac{2}{3}\times\left(\frac{11-8}{8\times11}+\frac{14-11}{11\times14}+...+\frac{29-26}{26\times29}\right)\)
\(=\frac{2}{3}\times\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{26}-\frac{1}{29}\right)\)
\(=\frac{2}{3}\times\left(\frac{1}{8}-\frac{1}{29}\right)\)
\(=\frac{7}{116}\)
VT
28 tháng 3 2022
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{20}\)
= \(\dfrac{9}{20}\)
DK
1
14 tháng 6 2023
A=1/5-1/8+1/8-1/11+...+1/602-1/605
=1/5-1/605
=24/121
SN
1
16 tháng 9 2023
a) \(2x^2-5x^2+6x+13=0\)
\(\Leftrightarrow-3x^2+6x+13=0\)
\(\Leftrightarrow3x^2-6x-13=0\left(1\right)\)
\(\Delta'=9+39=48>0\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{3}\)
Pt (1) có 2 nghiệm phân biệt là :
\(\left[{}\begin{matrix}x=\dfrac{3+4\sqrt[]{3}}{3}=1+\dfrac{4\sqrt[]{3}}{3}\\x=\dfrac{3-4\sqrt[]{3}}{3}=1-\dfrac{4\sqrt[]{3}}{3}\end{matrix}\right.\)
b) \(x^2-5x=-4\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
KK
4
F
15 tháng 7 2020
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}\)
\(=\frac{2004}{10045}\)
15 tháng 7 2020
Đề: Tính
\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)
Vậy \(A=\frac{2004}{10045}.\)
14 tháng 5 2018
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}\)
\(=\frac{3}{7}\)
14 tháng 5 2018
3/2x5 + 3/5x8 + 3/8x11 + 3/11x14
= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14
= 3/2 - 3/14
= 21/14 - 3/14
= 18/14
= 9/5
LN
0
7 tháng 10 2021
\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{98}{515}\)
\(\frac{339}{20}\)
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