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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) \(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{\left(3.4.....99\right).100}{2\left(3.4.....99\right)}\)
\(=\frac{100}{2}\)
\(=50\)
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
\(a,TH1:x-2021=0=>x=2021\)
\(Th2:x-2022=0=>x=2022\)
Vậy \(x\in\left\{2021;2022\right\}\)
\(b,x\left(8-5\right)=1080\)
\(x.3=1080\)
\(x=360\)
\(c,x^3=216< =>6^3=216=>x=3\)
\(d,5^5=3125\)
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
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
a) 27 . 75 + 25 .27 - 150 = 27 . (75 + 25) - 150
= 270 . 100 - 150
= 27 000 - 150
= 26 850
b) 3.52 - 16 : 22 = 12,25 - 16 : 4
= 12,25 - 4
= 8,25
c) 20 - [30 - (5 - 1)2 ] = 20 - [30 - 42 ]
= 20 - 30 - 16
= (-10) - 16
= -26
d) 60 : {[(12 - 3) . 2] + 2} = 60 : {[9 . 2] + 2}
= 60 : {18 + 2}
= 60 : 20
= 3
Theo đề ta có a=5k+2
b=5q+3
13a+11b=13(5k+2)+11(5q+3)=65k+26+55q+33=(65k+55q)+59
Ta có 65k+55q chia hết cho 5 vì mỗi số hạng đều chia hết cho 5
59 chia 5 dư 4
Vậy 13a+11b chia 5 dư 4
\(S=\frac{1}{2022}-\left(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\right)\)
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\)
\(A=\frac{5}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2022-2020}{2020.2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)\)
\(S=\frac{1}{2022}-A=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=-\frac{1262}{1011}\)