\(A=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{100}\right)\)

\(A=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{101}{100}\)

\(A=\frac{101}{2}\)

A = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{101}{100}\)

A = \(\frac{101}{2}\)

 \(A=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).....\left(1+\frac{1}{100}\right)\)

\(A=\frac{3}{2}.\frac{4}{3}.........\frac{101}{100}\)

\(A=\frac{101}{2}\)

a)

\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)

b)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\  = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)

\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)...\left(\frac{1}{999}+1\right)=\frac{3}{2}.\frac{4}{3}...\frac{1000}{999}=\frac{1000}{2}=500\)

biểu thức trên có giá trị là 500 nhé, mình chắc chắn đúng 100 % luôn , chúc bạn học tốt ^_^

500 chuan luon

a)\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right)..\left(\frac{1}{999}+1\right)=\frac{3}{2}.\frac{4}{3}....\frac{1000}{999}=\frac{3.4.5...1000}{2.3....999}=\frac{100}{2}=50\)

b)\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{1000}-1\right)=\left(-\frac{1}{2}\right).\left(\frac{-2}{3}\right)...\left(\frac{-999}{1000}\right)=-\frac{1}{1000}\)

\(D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{998}{999}.\frac{999}{1000}\)

\(D=\frac{1}{1000}\)( rút gọn những thừa số giống nhau ở tử và mẫu)

Vậy \(D=\frac{1}{1000}\)

D = \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{999}{1000}\)

D = \(\frac{1}{1000}\)

\(D=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{999}\right)\left(1-\frac{1}{1000}\right)\)

\(\Rightarrow D=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{998}{999}.\frac{999}{1000}\)

\(\Rightarrow D=\frac{1.2.3.....998.999}{2.3.4.....999.1000}\)

\(\Rightarrow D=\frac{1}{1000}\)

a) 2^3-(1/3)^0.9

=8-(1/3)^0

=8-1

=7

b) mk quên cách giải rồi

sorry mai nha

biết làm bài 1 thôi

\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)

= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)

lượt bỏ đi còn :

\(\frac{1000}{2}=500\)

Lời giải:

Ta có:
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{998}-1\right)\left(\frac{1}{999}-1\right)=\frac{1-2}{2}.\frac{1-3}{3}.....\frac{1-998}{998}.\frac{1-999}{999}\)

\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-997}{998}.\frac{-998}{999}\)

\(=\frac{(-1)(-2)(-3)....(-998)}{2.3.4...999}=\frac{1.2.3....998}{2.3.4...999}=\frac{1}{999}\)