Các câu hỏi tương tự
so sánh A=100+101 phần 101-100 và B=100^2+101^2 phần 101^2-100^2
A1+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}frac{A}{2}frac{1}{2}+frac{3}{2^4}+frac{4}{2^5}+....+frac{100}{2^{101}}A-frac{A}{2}left(1+frac{3}{2^3}+....+frac{100}{2^{100}}right)-left(frac{1}{2}+frac{3}{2^4}+.....+frac{100}{2^{101}}right)frac{A}{2}frac{1}{2}+frac{3}{2^3}+frac{1}{2^4}+frac{1}{2^5}+....+frac{1}{2^{100}}-frac{100}{2^{101}}frac{A}{2}frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+frac{1}{2^4}+....+frac{1}{2^{100}}-frac{1}{2^{101}}frac{A}{2}left(1-left(frac{1}{2}right)^{101}right).2-frac{10...
Đọc tiếp
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
Tính
\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{101\sqrt{100}+100\sqrt{101}}\)
So sánh A = 1 + 1/(√2) + 1/(√3) + ... + 1/(√100) và B = 2√(101) - 1
Tính \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{100}+100\sqrt{101}}\)
Giai hpt \(\hept{\begin{cases}\frac{101}{x}+\frac{100}{y}=A-100\\x+\frac{7200y}{101}=A\end{cases}}\left(A\inℕ^∗\right)\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
Tính \(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right)\div\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)-2\)
So sánh: \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{100}+\sqrt{101}}\) với \(B=\frac{181}{20}\)