Lời giải:

$BCNN(m,n) = 2^4.3^5.5.7^6.11^4$

\(\overline{ab}+\overline{ba}=187\)

\(\Rightarrow10a+b+10b+a=187\)

\(\Rightarrow11a+11b=187\)

\(\Rightarrow11\left(a+b\right)=187\)

\(\Rightarrow a+b=187:11=17\)

Mà \(0< a,b\le9\) nên \(\overline{ab}\in\left\{98;89\right\}\)

Ko có số nào có 2 chữ số chia hết cho 1 000 000 000 cả 

A = \(\dfrac{8}{9}+\dfrac{24}{25}\) + \(\dfrac{48}{49}\) + ... + \(\dfrac{10200}{10201}\)

A = \(\dfrac{8}{3^2}\) + \(\dfrac{24}{5^2}\) + \(\dfrac{48}{7^2}\)  + ... + \(\dfrac{10200}{101^2}\)

Xét dãy số: 3; 5; 7;...; 101

Dãy số trên là dãy số cách đều với khoảng là: 5 - 3 = 2

Số số hạng của dãy số trên là: (101 - 3): 2 + 1 = 50

Vậy A có 50 hạng tử

     \(\dfrac{8}{9}\)  < \(1\)

     \(\dfrac{24}{25}\) < 1

    \(\dfrac{48}{49}\) < 1

  ..................

\(\dfrac{10200}{10201}\) < 1

Cộng vế với vế ta có: 

A = \(\dfrac{8}{9}\) + \(\dfrac{24}{25}\) + \(\dfrac{48}{49}\) +....+ \(\dfrac{10200}{10201}\) < 1 x 50

A < 50 < 99,75 (trái với đề bài)

Vậy việc chứng minh A > 99,75 là điều không thể xảy ra. 

\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)

\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)

\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1011}{4048}\)

\(A=\frac{1}{2.6}+\frac{1}{3.8}+\frac{1}{4.10}+...+\frac{1}{2023.4048}\\=\frac12\left(\frac{2}{2.6}+\frac{2}{3.8}+\frac{2}{4.10}+...+\frac{2}{2023.4048}\right)\\=\frac12\left( \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2023.2024}\right)\\=\frac12\left(\frac12-\frac13+\frac13-\frac14+\frac14-\frac15+...+\frac{1}{2023}-\frac{1}{2024}\right)\\=\frac12\left(\frac12-\frac{1}{2024}\right) \\=\frac12.\frac{1011}{2024}=\frac{1011}{4048}\)

\(\overline{1ab}+36=\overline{ab1}\)

=>\(100+\overline{ab}+36=10\overline{ab}+1\)

=>\(9\overline{ab}=100+36-1=135\)

=>\(\overline{ab}=15\)

=>a=1; b=5

45 - 8 = 12 ở đâu ra vậy em?

Thay x = 5 ý ạ

\(5:\dfrac{3}{14}-4\dfrac{4}{5}:\dfrac{3}{4}\\ =5.\dfrac{14}{3}-\dfrac{24}{5}.\dfrac{4}{3}\\ =\dfrac{70}{3}-\dfrac{32}{5}\\ =\dfrac{254}{15}\)

.

\(-\dfrac{13}{8}.\left(\dfrac{8}{13}+\dfrac{32}{38}\right)-\dfrac{15}{7}\\ =-\dfrac{13}{8}.\left(\dfrac{8}{13}+\dfrac{16}{19}\right)-\dfrac{15}{7}\\ =-\dfrac{13}{8}.\dfrac{360}{247}-\dfrac{15}{7}\\ =-\dfrac{45}{19}-\dfrac{15}{7}\\ =-\dfrac{600}{133}\)

\(\dfrac{x}{3}-\dfrac{1}{8}+\dfrac{4}{2}-\dfrac{3}{8}=\dfrac{5}{50}\\ \dfrac{x}{3}-\dfrac{4}{8}+2=\dfrac{1}{10}\\ \dfrac{x}{3}-\dfrac{1}{2}+2=\dfrac{1}{10}\\ \dfrac{x}{3}+\dfrac{3}{2}=\dfrac{1}{10}\\ \dfrac{x}{3}=\dfrac{1}{10}-\dfrac{3}{2}\\ \dfrac{x}{3}=-\dfrac{7}{5}\\ x=-\dfrac{7}{5}.3\\ x=-\dfrac{21}{5}\)

\(A=\dfrac{-3}{5}+\left(\dfrac{-2}{5}-99\right)\)

\(A=\dfrac{-3}{5}+\dfrac{-2}{5}-99\)

\(A=\left(\dfrac{-3}{5}+\dfrac{-2}{5}\right)-99\)

\(A=\dfrac{-5}{5}-99\)

\(A=-1-99\)

\(A=-100\)