Thu gọn :
A = \(\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)
1.\(\dfrac{1}{2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}+\dfrac{7}{22.29}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{29}\)
=1-1/29
=28/29
\(\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+...+\dfrac{14}{92.106}\)
\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{106}\)
\(=\dfrac{1}{4}-\dfrac{1}{106}=\dfrac{51}{212}\)
\(\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+...+\dfrac{14}{92.106}\)
= \(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{92}-\dfrac{1}{106}\)
= \(\dfrac{1}{4}-\dfrac{1}{106}\)
= \(\dfrac{51}{212}\)
Tính
N = \(\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)
\(N=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\\ N=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{46}-\dfrac{1}{56}\\ N=1-\dfrac{1}{56}\\ N=\dfrac{55}{56}\)
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
Thu gọn :
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Biết nhưng ko giúp đâu, bài dễ thì phải động não nha
Bài này có lẽ không làm được bn ạ, vì ở dưới mẫu có 2 số hạng \(\dfrac{2}{3}\)
chia cho 0 rùi, mà không có số nào chia được cho 0 cả, lặp luôn cả cái này ở tử nữa!!!
6. Tính
\(A=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{31.34}\)
\(B=1-5+5^2-5^3+5^4-...-5^{39}\)
a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)
Tính:
a) A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{998.999}\) + \(\dfrac{1}{999.1000}\)
b) B = \(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\) + \(\dfrac{1}{11.16}\) +...+ \(\dfrac{1}{495.500}\)
c) C = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) +...+ \(\dfrac{1}{998.999.1000}\)
(Mong mn giúp ạ)
a.
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=1-\frac{1}{1000}=\frac{999}{1000}$
b.
$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$
$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$
$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$
$=1-\frac{1}{500}=\frac{499}{500}$
$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$
c.
$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$
$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$
$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$
1. Một người đi xe máy từ A đến B với vận tốc trung bình \(36\dfrac{1}{4}\)km/h hết 3,2h. Lúc về người ấy đi với vận tốc trung bình 40 km/h. Tính thời gian người ấy đi từ A đến B.
2.
a)M=\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
b)N=\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{197..199}\)
c)P=\(\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...+\dfrac{10}{46.56}\)
2)
a) M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)
1) Quãng đường AB là :
\(36\dfrac{1}{4}\)km/h . 3.2h = 116 ( km )
Thời gian người ấy đi từ A đến B lúc về là :
116 : 40 = 2.9 ( giờ )
Đ/S : 2.9 giờ
N = \(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{197.199}\)
N = \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{197.199}\right)\)
N = \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
N = \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{199}\right)\)
N = \(\dfrac{3}{2}.\dfrac{194}{995}\)
N = \(\dfrac{291}{995}\)
Bài 1 : Tính
a) A = \(\left(\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(\dfrac{55}{123}+\dfrac{555}{1234}+\dfrac{5555}{12345}\right)\)
b) B = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{101.106}\)
c) C = \(\dfrac{2x^2+3x-1}{3x-2}\) với \(\left|x-1\right|=2\)
a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)