a) Ta thấy \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};...;\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

b) \(A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(A.B=\frac{1.\left(3.5...99\right).\left(2.4.6...100\right)}{\left(2.4.6...100\right).\left(3.5.7...99\right).101}=\frac{1}{101}\)

c) vì A < b nên A . A < A . B < \(\frac{1}{101}< \frac{1}{100}\)

do đó : A . A  < \(\frac{1}{10}.\frac{1}{10}\)suy ra A < \(\frac{1}{10}\)

a³ +b³ = (a+b)(a² -ab + b²) = 3(a² +b² - (-10)) (1)

mà a² + b² = (a+b)² - 2ab = 3² + 2.10 = 29 (2)

thay(1) vảo (2) có: A = 3(29+10) = 127

1:

a) abcabc = abc x 1001

Mà 1001 chia hết cho 11 nên abc x 1001 chia hết cho 11. Suy ra abcabc chia hết cho 11.

b) \(1-\left(5\frac{4}{9}+x+7\frac{7}{18}\right):15\frac{3}{4}=0\)

\(1-\left(\frac{49}{9}+x+\frac{133}{18}\right)=0\)

\(\frac{231}{18}+x=1-0=1\)

\(x=1-\frac{231}{18}=\frac{-213}{18}\)

Ko bt câu b có đúng ko thử lại nha.

Từ \(a+b=10=>\left(a+b\right)^2=100=>a^2+2ab+b^2=100=>a^2+2.4+b^2=100.\)

\(\Rightarrow a^2+b^2=92\)

\(\left(a^2+b^2\right).\left(a^3+b^3\right)=a^5+a^2b^3+a^3b^2+b^5=92.880\) 

\(=>a^5+b^5+a^2b^2\left(a+b\right)=80960\) 

\(=>a^5+b^5+\left(ab\right)^2\left(a+b\right)=80960\)

\(=>a^5+b^5+4^2.10=80960\)

\(=>a^5+b^5=80800\)

\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)

\(=>A=\frac{1\cdot2+4\cdot1\cdot2+9\cdot1\cdot2+16\cdot1\cdot2+25\cdot1\cdot2}{3\cdot4+4\cdot3\cdot4+9\cdot3\cdot4+16\cdot3\cdot4+25\cdot3\cdot4}\)

\(=>A=\frac{\left(1+4+9+16+25\right)\cdot1\cdot2}{\left(1+4+9+16+25\right)\cdot3\cdot4}=\frac{1}{6}=\frac{111111}{666666}\)

Mà \(\frac{111111}{666666}< \frac{111111}{666665}\)

\(=>A< B\)