1+1 = 1+1

2

\(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+98\right)}{1.98+2.97+3.96+...+97.2+98.1}\)
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.98+2.97+3.96+...+98.1}=1\)



 

\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)...\left(1-\frac{3}{97}\right)\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.....\frac{94}{97}.\frac{97}{100}\)

\(=\frac{1.4.7.....94.97}{4.7.10.....97.100}\)

\(=\frac{1}{100}\)

a)   \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

\(=\frac{1}{5}\)

b)   \(\left(1-\frac{3}{4}\right).\left(1-\frac{3}{7}\right).\left(1-\frac{3}{10}\right)........\left(1-\frac{3}{97}\right).\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.......\frac{94}{97}.\frac{97}{100}\)

\(=\frac{1}{100}\)

a: \(D=\frac{10}{100}+\frac{10}{150}+\frac{10}{210}+\cdots+\frac{10}{1200}\)

\(=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\cdots+\frac{1}{120}\)

\(=\frac{2}{20}+\frac{2}{30}+\cdots+\frac{2}{240}=2\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\cdots+\frac{1}{15\cdot16}\right)\)

\(=2\left(\frac14-\frac15+\frac15-\frac16+\cdots+\frac{1}{15}-\frac{1}{16}\right)=2\left(\frac14-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac38\)

b: \(E=1\cdot2+2\cdot3+\cdots+99\cdot100\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)

\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)

\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\left(99+1\right)}{2}=\frac{99\cdot100\cdot199}{6}+99\cdot\frac{100}{2}\)

\(=33\cdot50\cdot199+99\cdot50\)

\(=33\cdot50\cdot\left(199+3\right)=33\cdot50\cdot202=33\cdot101\cdot100=333300\)

c: \(F=1^2+2^2+\cdots+98^2\)

\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}=\frac{98\cdot99\cdot197}{6}=49\cdot33\cdot197=318549\)

1: Ta có: \(20-2\left(x+4\right)=4\)

\(\Leftrightarrow2\left(x+4\right)=16\)

\(\Leftrightarrow x+4=8\)

hay x=4

5: Ta có: \(\left(x+1\right)^3=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

Ta có:A= \(\frac{10^{97}+1}{10^{98}+1}=10\cdot\left(\frac{10^{97}+1}{10^{98}+1}\right)=\frac{10^{98}+10}{10^{98}+1}\)=\(\frac{10^{98}+1+9}{10^{98}+1}\)=\(\frac{9}{10^{98}+1}+1\)

B=\(\frac{10^{96}+1}{10^{97}+1}=10\cdot\left(\frac{10^{96}+1}{10^{97}+1}\right)=\frac{10^{97}+10}{10^{97}+1}\)=\(\frac{10^{97}+1+9}{10^{97}+1}\)=\(\frac{9}{10^{97}+1}+1\)

Vì  \(\frac{9}{10^{98}+1}+1\)< \(\frac{9}{10^{97}+1}+1\)\(\left(10^{98}+1>10^{97}+1\right)\)

Nên A<B

A = 1 - 2 - 3 - 4 + 5 - 6 -  7 - 8 + ........... + 97 - 98 - 99 - 100 (100 số )

A = (1 - 2 - 3 - 4) + (5 - 6 - 7 - 8) + ................ + (97 - 98 - 99 - 100)

(25 cặp , tính bằng cách lấy số cả dãy chia cho số số của mỗi cặp )

A = (-8) . 25 

A = -200

59 nha