1. Có 99 chữ số từ 24,01 - 24,99

2. Một người đi quãng đường AB vận tốc 15 km trên nửa quãng đường đầu và vận tốc 10 km trên nửa quãng đường sau. Vận tốc trung bình của người đó trên cả quãng đường AB là 12,5 km h.

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=2021\)

\(\left(x+x+x+...+x\right)+\left(1+2+...+99\right)=2021\)

\(100x+\left(1+2+...+99\right)=2021\)

Ta tính tổng \(A=1+2+...+99\)  (Số số hạng của tổng là 99)

\(A=\left(1+99\right)+\left(2+98\right)+...+\left(49+51\right)+50\)

\(A=100+100+...+100+50=100\times49+50=4950\)

Vậy \(100x+4950=2021\)

Suy ra \(100x=2021-4950=-2929\), hay \(x=-29,29\)

a) \(\dfrac{9}{11}-\dfrac{7}{8}+\dfrac{13}{11}+\dfrac{-1}{8}=\left(\dfrac{9}{11}+\dfrac{13}{11}\right)-\left(\dfrac{7}{8}+\dfrac{1}{8}\right)=2-1=1\)

b) \(\dfrac{1}{9}+\dfrac{4}{9}:2^2-5.\dfrac{6}{5}=\dfrac{1}{9}+\dfrac{4}{9}:4-6=\dfrac{1}{9}+\dfrac{1}{9}-6=\dfrac{2}{9}-6=-\dfrac{52}{9}\) (Đề câu này không được rõ, em kiểm tra lại nhé).

c) \(\dfrac{7}{12}.\dfrac{2}{3}-\dfrac{5}{3}.\dfrac{7}{12}+\dfrac{7}{12}.3=\dfrac{7}{12}\left(\dfrac{2}{3}-\dfrac{5}{3}+3\right)=\dfrac{7}{12}.\left(-1+3\right)=\dfrac{7}{12}.2=\dfrac{7}{6}\)

d) \(\left(2,3-53,3\right)-\left(46,7-7,7\right)=2,3-53,3-46,7+7,7=\left(2,3+7,7\right)-\left(53,3+46,7\right)=10-100=-90\)

\(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\\ =4,35-2,67+1,65+3,54-6,33\\ =\left(4,35+1,65\right)-\left(2,67+6,33\right)+3,54\\ =6-9+3,54=3,54-3=0,54\)

4,35−(2,67−1,65)+(3,54−6,33)

=4,35−2,67+1,65+3,54−6,33

=(4,35+1,65)−(2,67+6,33)+3,54

=6−9+3,54

=3,54−3

=0,54

a) \(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\)

\(=4,35-2,67+1,65+3,54-6,33\\ =\left(4,35+1,65\right)-\left(2,67+6,33\right)+3,54\\ =6-9+3,54=3,54-3=0,54\)

b) \(\left(-0,5\right).\left(-0,5\right).\left(-0,8\right)\)

\(=\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{4}{5}\right)\\ =-\left(\dfrac{1.1.4}{2.2.5}\right)=-\dfrac{1}{5}\)

c) \(3,58.24,45+3,58.75,55\)

\(=3,58\left(24,45+75,55\right)\\ =3,58.100=358\)

d) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right).\dfrac{5}{7}\\ =\dfrac{4.7.3}{3.5.4}-\dfrac{4}{5}.\dfrac{5}{7}\\ =\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{29}{35}\)

e) \(\left(3,6-2\dfrac{2}{5}\right).\dfrac{-5}{3}+3\left(2\dfrac{1}{2}:50\%\right)\)

\(=\left(\dfrac{18}{5}-\dfrac{12}{5}\right).\dfrac{-5}{3}+3.\left(\dfrac{5}{2}:\dfrac{1}{2}\right)\\ =\dfrac{6}{5}.\dfrac{-5}{3}+3.5=-2+15=13\)

`Answer:`

\(x+\frac{1}{4}=-\frac{3}{4}.\frac{23}{-15}.\left(-\frac{45}{92}\right)\) (Mình sửa đề nhé.)

\(\Leftrightarrow x+\frac{1}{4}=-\frac{3}{4}.\left(-\frac{23}{15}\right).\left(-\frac{45}{92}\right)\)

\(\Leftrightarrow x+\frac{1}{4}=-\frac{3}{4}.\frac{3}{4}\)

\(\Leftrightarrow x+\frac{1}{4}=-\frac{9}{16}\)

\(\Leftrightarrow x=-\frac{9}{16}-\frac{1}{4}\)

\(\Leftrightarrow x=-\frac{13}{16}\)

`Answer:`

\(5.\left(x-2\frac{1}{5}\right)^2-\frac{1}{5}=9\frac{3}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2-\frac{1}{5}=\frac{48}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{48}{5}+\frac{1}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{49}{5}\)

\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\frac{49}{5}:5\)

\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\left(\frac{7}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{11}{5}=\frac{7}{5}\\x-\frac{11}{5}=-\frac{7}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{5}+\frac{11}{5}\\x=-\frac{7}{5}+\frac{11}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{18}{5}\\x=\frac{4}{5}\end{cases}}\)