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

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)

\(\Leftrightarrow x=122\)

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, \(=\frac{49}{40}=1,225\)

\(a,TH1:x-2021=0=>x=2021\)

\(Th2:x-2022=0=>x=2022\)

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

a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)

\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)

\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)

b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)

\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)

\(=2\left(1-\dfrac{1}{199}\right)\)

\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)

giúp mk với nha

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

3/8*5/4=15/32

4/7*2/13=8/91

\(\frac{3}{8}\)* \(\frac{5}{4}\)= \(\frac{15}{32}\)                                   \(\frac{4}{7}\)* \(\frac{2}{13}\)= \(\frac{8}{91}\)

Bạn chỉ cần lấy tử số nhân tử số , mẫu số nhân mẫu số thôi

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)