Ta có:

\(A=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(2A=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)

\(2A+A=\left(1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots+\frac{1}{2^{98}}-\frac{1}{2^{99}}\right)+\left(\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots+\frac{1}{2^{99}}-\frac{1}{2^{100}}\right)\)

\(3A=1-\frac{1}{2^{100}}\)

\(A=\left(1-\frac{1}{2^{100}}\right):3=\frac13-\frac{1}{3.2^{100}}\)

làm tròn đến hàng nào vậy bn?

0

Ta có:

\(\frac{11}{-9}=\frac{11\cdot\left(-7\right)}{-9\cdot\left(-7\right)}=\frac{-77}{63}\)

\(\frac{-9}{7}=\frac{-9\cdot9}{7\cdot9}=\frac{-81}{63}\)

Vì \(-77>-81\) nên \(\frac{-77}{63}>\frac{-81}{63}\)

Vậy \(\frac{11}{-9}>\frac{-9}{7}\)

Cách viết cho ta được phân số là: \(\frac{-0}{2024}\)

\(35,5dm^3=0,0335m^3\)

\(35000dm^3=35m^3\)

3,749

A