15: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(\dfrac{1}{x+1}=\dfrac{1}{2022}\)

=>x+1=2022

=>x=2021

14: \(3^{x+1}+3^{x+1}\cdot4=45\)

=>\(3^x\cdot3+3^x\cdot12=45\)

=>\(3^x=3\)

=>x=1

11: \(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\)

=>\(\dfrac{7}{12}\left(x+\dfrac{13}{21}\right)=\dfrac{13}{15}-\dfrac{7}{10}=\dfrac{1}{6}\)

=>\(x+\dfrac{13}{21}=\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{1}{6}\cdot\dfrac{12}{7}=\dfrac{2}{7}\)

=>\(x=\dfrac{2}{7}-\dfrac{13}{21}=-\dfrac{7}{21}=-\dfrac{1}{3}\)

12: \(720:\left[41-\left(2x-5\right)\right]=2^2\cdot5\)

=>\(41-\left(2x-5\right)=\dfrac{720}{20}=36\)

=>2x-5=5

=>2x=10

=>x=5

9: \(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}\)

=>\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{4}{7}\)

=>\(\left\{{}\begin{matrix}x=7\\y=21\cdot\dfrac{4}{7}=12\end{matrix}\right.\)

\(\left(\dfrac{31}{42}\times\dfrac{17}{25}+\dfrac{31}{42}\right)\times\left(-5\right)^2\)

\(=\dfrac{31}{42}\times\left(\dfrac{17}{25}+1\right)\times25\)

\(=\dfrac{31}{42}\times\dfrac{42}{25}\times25=31\)

d: \(\left(\dfrac{2}{5}-\dfrac{2}{3}\right)-\dfrac{7}{5}=\dfrac{2}{5}-\dfrac{7}{5}-\dfrac{2}{3}=-1-\dfrac{2}{3}=-\dfrac{5}{3}\)

b: \(\dfrac{14}{13}+\left(-\dfrac{1}{13}-\dfrac{19}{20}\right)=\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)

f: \(\dfrac{7}{8}-\dfrac{3}{8}\left(1-\dfrac{2}{3}\right)\)

\(=\dfrac{7}{8}-\dfrac{3}{8}\cdot\dfrac{1}{3}\)

\(=\dfrac{7}{8}-\dfrac{1}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)

\(d,\left(\dfrac{2}{5}-\dfrac{2}{3}\right)-\dfrac{7}{5}\)

\(=\dfrac{2}{5}-\dfrac{2}{3}-\dfrac{7}{5}\)

\(=\left(\dfrac{2}{5}-\dfrac{7}{5}\right)-\dfrac{2}{3}\)

\(=-1-\dfrac{2}{3}\)

\(=-\dfrac{3}{3}-\dfrac{2}{3}=-\dfrac{5}{3}\)

\(e,\dfrac{14}{13}+\left(-\dfrac{1}{13}-\dfrac{19}{20}\right)\)

\(=\dfrac{14}{13}+-\dfrac{1}{13}-\dfrac{19}{20}\)

\(=\left(\dfrac{14}{13}+-\dfrac{1}{13}\right)-\dfrac{19}{20}\)

\(=1-\dfrac{19}{20}\)

\(=\dfrac{1}{20}\)

\(f,\dfrac{7}{8}-\dfrac{3}{8}.\left(1-\dfrac{2}{3}\right)\)

\(=\dfrac{7}{8}-\dfrac{3}{8}.\dfrac{1}{3}\)

\(=\dfrac{7}{8}-\dfrac{1}{8}.1\)

\(=\dfrac{6}{8}=\dfrac{3}{4}\)

\(\left(-0,6\right).\left(-0,8\right).\left(-0,6\right)\)

\(=\left(-0,6\right).\left(-0,8\right).\left(-0,6\right).1\)

\(=\left(-0,6\right).\left(-0,8.1\right)\)

\(=\left(-0,6\right).\left(-0,8\right)\)

\(=0,48\)

\(S=\dfrac{1}{2^1}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2002}{2^{2002}}+\dfrac{2003}{2^{2003}}\)

\(\Rightarrow2S=1+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2002}{2^{2001}}+\dfrac{2003}{2^{2002}}\)

Trừ vế cho vế:

\(\Rightarrow2S-S=1+\left(\dfrac{2}{2^1}-\dfrac{1}{2^1}\right)+\left(\dfrac{3}{2^2}-\dfrac{2}{2^2}\right)+...+\left(\dfrac{2003}{2^{2002}}-\dfrac{2002}{2^{2002}}\right)-\dfrac{2003}{2^{2003}}\)

\(\Rightarrow S=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2002}}-\dfrac{2003}{2^{2003}}\)

\(\Rightarrow2S=2+1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2001}}-\dfrac{2003}{2^{2002}}\)

Trừ vế cho vế:

\(\Rightarrow2S-S=2-\dfrac{2004}{2^{2002}}+\dfrac{2003}{2^{2003}}\)

\(\Rightarrow S=2-\dfrac{1}{2^{2003}}\left(2004.2-2003\right)\)

\(\Rightarrow S=2-\dfrac{2005}{2^{2003}}< 2\)

\(A=\dfrac{2023}{1\cdot2}+\dfrac{2023}{2\cdot3}+...+\dfrac{2023}{2022\cdot2023}\)

\(=2023\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=2023\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=2023\left(1-\dfrac{1}{2023}\right)=2023\cdot\dfrac{2022}{2023}=2022\)

\(A=\dfrac{2023}{1.2}+\dfrac{2023}{2.3}+\dfrac{2023}{3.4}+...+\dfrac{2023}{2022.2023}\)

\(A=\dfrac{2023}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\right)\)

\(A=\dfrac{2023}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(A=\dfrac{2023}{1}.\left(1-\dfrac{1}{2023}\right)\)

\(A=\dfrac{2023}{1}.\dfrac{2022}{2023}\)

\(A=1.2022\)

\(A=2022\)

Vậy \(A=2022\)

b: Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)

=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-...-\dfrac{1}{2^{100}}\)

=>\(A=1-\dfrac{1}{2^{100}}< 1\)

2A ạ

Bài 8:

Số học sinh đi xe đạp là:

\(45\cdot\dfrac{4}{9}=20\left(bạn\right)\)

Số học sinh đi xe buýt là:

\(12:\dfrac{3}{4}=12\cdot\dfrac{4}{3}=16\left(bạn\right)\)

Số học sinh đi bộ là 45-20-16=45-36=9(bạn)

Số học sinh đi xe đạp là:

\(45\times\dfrac{4}{9}=20\) (học sinh)

Số học sinh đi xe buýt là:

\(12:\dfrac{3}{4}=16\) (học sinh)

Số học sinh đi bộ là:

\(45-\left(20+16\right)=9\) (học sinh)

                                               Giải

Số lít không khí 1 người thở trong 1 phút là

15. 0,53= 7,95[lít]

Số lít không khí 1 người thở trong 1 giờ là 

7,95.60 = 477 [lít]

Số lít không khí 6 người thở trong 1 giờ là 

477.6= 2862 [lít]

Số kg không khí 6 người thở trong 1 giờ là

2862.0,032 = 91,584[lít]

                    xấp xỉ   91,6 lít

                    ĐS [ TỰ GHI NHEN]

a: Trên tia Ox, ta có: OB<OA

nên B nằm giữa O và A

=>OB+BA=OA

=>BA+4=10

=>BA=6(cm)

M là trung điểm của BA

=>\(MB=MA=\dfrac{BA}{2}=3\left(cm\right)\)

Vì M nằm giữa B và A

và B nằm giữa O và A

nên B nằm giữa O và M

=>OM=OB+BM=4+3=7(cm)

b: B là trung điểm của MN

=>\(MN=2\cdot MB=2\cdot3=6\left(cm\right)\)

Vì MN<MO

nên N nằm giữa M và O

=>MN+NO=MO

=>NO+6=7

=>NO=1(cm)