a: \(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)

\(=\left(4\cdot32\right):\left(\frac{8}{16}\right)\)

\(=128:\frac12=128\cdot2=256\)

b: \(B=3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)

\(=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=\frac{3^8}{3^8}\cdot3^2=3^2=9\)

c: \(D=\left\lbrace\left(0,1\right)^2\right\rbrace^0+\left\lbrack\left(\frac17\right)^1\right\rbrack^2:\frac{1}{49}\cdot\left\lbrack\left(2^2\right)^3:2^5\right\rbrack\)

\(=1+\left(\frac17\right)^2\cdot49\cdot2^6:2^5\)

\(=1+49\cdot\frac{1}{49}\cdot2=1+2=3\)

d: \(C=\left(-0,5\right)^5:\left(-0,5\right)^3-\left(\frac{17}{2}\right)^7:\left(\frac{17}{2}\right)^6\)

\(=\left(-0,5\right)^{5-3}-\left(\frac{17}{2}\right)\)

\(=\left(-0,5\right)^2-\frac{17}{2}=0,25-\frac{17}{2}=\frac14-\frac{34}{4}=-\frac{33}{4}\)

\(A=-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}\)

\(\Rightarrow-A=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{10-9}{9.10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\Rightarrow A=-\dfrac{2}{5}\)

a) Đặt biểu thức trên là A

\(A=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}+\frac{2}{11}-\frac{5}{7}+\frac{3}{7}-\frac{1}{5}\)

\(A=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{-2}{11}+\frac{2}{11}\right)+\frac{5}{9}+\frac{7}{13}-\frac{5}{7}\)

\(A=0+0+0+\frac{5}{9}+\frac{7}{13}-\frac{5}{7}\)

\(A=\frac{128}{117}-\frac{5}{7}\)

\(A=\frac{311}{819}\)

1/99-1/99.98-1/98.97-.....-1/2.1

= 1/99-(1/99.98+1/98.97+......+1/2.1

=1/99-(1/99-1)

=1

a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(A=1-\frac{1}{99}\)

\(A=\frac{98}{99}\)

thay A vào, ta được :

\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)

b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)

\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)

đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)

\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)

\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(A=2.\left(1-\frac{1}{99}\right)\)

\(A=2.\frac{98}{99}\)

\(A=\frac{196}{99}\)

Thay A vào, ta được :

\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Phạm Trần Khánh An : l.i.k.e tiếp cái con khỉ