Mình gửi lại câu hỏi nhé:
S = 1/2022 - 5/2.4 - 5/4.6 - ... - 5/2020.2022
(Dấu (.) là dấu nhân các bạn nhé)).
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Những câu hỏi liên quan
16 tháng 4 2017
a) B = 5/16 : 0,125 - ( 9/4 - 0,6 ) * 10/11
B = 5/16 * 8 - 9/4 * 10/11 + 0,6 * 10/11
B = 5/2 - 45/22 + 3/11
B = 55/22 - 45/22 + 6/22
B = 8/11
b) \(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{\left(3.4.....99\right).100}{2\left(3.4.....99\right)}\)
\(=\frac{100}{2}\)
\(=50\)
16 tháng 4 2017
a)B= 5/6 : 1/8 - (9/4 - 3/5 ) * 10/11
= 5/2 - 33/20 * 10/11
= 5/2 - 3/2
= 1
b) 3/2.4/3.5/4....100/99
= 3.4.5...100/2.3.4...99
=100/2
=50
MN
7 tháng 6 2021
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)
\(\Leftrightarrow x=122\)
11 tháng 4 2017
a, 32,5x6,5+3,5x32,5
=32,5x(6,5+3,5)
=32,5x10
=325
b, =48788,5
c, \(=\frac{29}{20}=1,45\)
d, \(=\frac{49}{40}=1,225\)
NH
22 tháng 9 2021
\(a,TH1:x-2021=0=>x=2021\)
\(Th2:x-2022=0=>x=2022\)
Vậy \(x\in\left\{2021;2022\right\}\)
\(b,x\left(8-5\right)=1080\)
\(x.3=1080\)
\(x=360\)
\(c,x^3=216< =>6^3=216=>x=3\)
\(d,5^5=3125\)
22 tháng 9 2021
a) ( x- 2021) * ( x- 2022) = 0
=> \(\orbr{\begin{cases}x-2021=0\\x-2022=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2021\\x=2022\end{cases}}}\)
b) b. 8x - 5x = 2022
=> 3x = 2022
=> x = 674
c) \(5\cdot x^3=1080\)
=> \(x^3=216\)
=> \(x^3=6^3\)
=> x = 6
d) \(5^x=3125\)
=> \(5^x=5^5\)
=> x = 5
RT
2
11 tháng 7 2021
a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)
\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)
\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)
11 tháng 7 2021
b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
\(=2\left(1-\dfrac{1}{199}\right)\)
\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)
9 tháng 4 2017
3/8 x 5/4 =15/32
4/7 x 2/13=8/91
cách làm là lấy tử nhân tử mẫu nhân mẫu
cái này học từ lớp 4 rồi còn gì
NT
1
29 tháng 4 2021
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
\(S=\frac{1}{2022}-\left(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\right)\)
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\)
\(A=\frac{5}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2022-2020}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)\)
\(S=\frac{1}{2022}-A=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=-\frac{1262}{1011}\)