b: \(B=\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{100}}\)
=>\(2B=1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots-\frac{1}{2^{99}}\)
=>2B+B=\(1-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\ldots-\frac{1}{2^{99}}+\frac12-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\cdots-\frac{1}{2^{100}}\)
=>3B=\(1-\frac{1}{2^{100}}\)
=>\(B=\frac13\left(1-\frac{1}{2^{100}}\right)\)
a: \(\frac{0,4-\frac29+\frac{2}{11}}{1,4-\frac79+\frac{7}{11}}-\frac{\frac13-0,25+\frac15}{1\frac16-0,875+0,7}\)
\(=\frac{\frac25-\frac29+\frac{2}{11}}{\frac75-\frac79+\frac{7}{11}}-\frac{\frac13-\frac14+\frac15}{\frac76-\frac78+\frac{7}{10}}\)
\(=\frac{2\left(\frac15-\frac19+\frac{1}{11}\right)}{7\left(\frac15-\frac19+\frac{1}{11}\right)}-\frac{\frac13-\frac14+\frac15}{\frac72\left(\frac13-\frac14+\frac15\right)}=\frac27-1:\frac72=\frac27-\frac27=0\)
TA có: \(A=\frac{2018}{2019}\cdot\left(\frac{0,4-\frac29+\frac{2}{11}}{1,4-\frac79+\frac{7}{11}}-\frac{\frac13-0,25+\frac15}{1\frac16-0,875+0,7}\right)\)
\(=\frac{2018}{2019}\cdot0=0\)
làm giúp mình bài 1 với, mình cần gấp ( nếu rảnh làm hộ mình bài 2 cũng dc )
