1\(\frac{1}{12}-\frac{1}{20}-...-\frac{1}{380}=?\)
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Đặt bt trên là A nha
Đổi |x-1|=|1-x|
Suy ra A=|1-x|+x-2|+|x-3|
Áp dụng BĐTGTTĐ ta có
A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra
Vậy x =2
b,
\(\left|3x+\frac{1}{2}\right|\ge0\)
\(\left|3x+\frac{1}{6}\right|\ge0\)
..........
\(\left|3x+380\right|\ge0\)
Suy ra đề bài \(\ge\)0
suy ra 58x \(\ge\)0
Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)
Tự tính nhé hok tốt
1/2.3+1/3.4+1/4.5+...+1/380
=1/2.3+1/3.4+1/4.5+...+1/19.20
=1/2-1/3+1/3-1/4+1/4-1/5+...+1/19-1/20
=1/2-1/20
=9/20
tick nha bn cám ơn
A=3.(1/2 +1/6 +1/12 +1/20+...+1/380)
A:3=1/1.2+1/2.3+1/3.4 +...+1/19.20
A:3=1-1/20
A:3=19/20
A=19/20.3
A=57/20
A=3/1X2+3/2X3+3/3X4+3/4X5+...+3/19X20
A=3(1/1X2+1/2X3+1/3X4+1/4X5+..+1/19X20
A=3(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/19-1/20)
A=3(1/1-1/20)
A=3.19/20
A=57/20
tk cho mk nha
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
=( 1/12+9/12+2/12+1/12) + ( 2/7+5/7) + (10/20+9/20)
= 13/12 +1+19/20
đến đây bạn tự tính nha
Ta xét : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)
\(=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\)
Vì \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
nên \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\) ( đpcm )
\(\frac{1}{12}-\frac{1}{20}-...-\frac{1}{380}\)
\(=\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{19.20}\)
\(=\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{6}\right)-...-\left(\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-...-\frac{1}{19}+\frac{1}{20}\)
\(=\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{20}\)
\(=-\frac{7}{60}\)