Bài 5 :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{59}\)
\(A=1-\frac{1}{50}\)
từ trên ta có : \(1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)
\(\frac{2017}{1\times2\text{×}3}+\frac{2017}{2\text{×}3\text{×}4}+\frac{2017}{3\text{×}4\text{×}5}+..+\frac{2017}{19\text{×}20\text{×}21}\)
Cho \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{\text{n}}{5^{\text{n}-2}}+...+\frac{11}{5^{12}}\) với \(\text{n}\in\text{N }\).CMR:\(A< \frac{1}{16}\)
bai 1 \(\frac{-3}{\text{2}}+\frac{5}{7}+\frac{-31}{14}< hoac=\text{x}< \frac{1}{\text{2}}+\frac{1}{3}+\frac{1}{6}\)\(\frac{1}{6}\)
bai 2 \(\frac{\text{x}+4}{\text{x}-\text{2}}+\frac{\text{2}\text{x}-5}{\text{x}-\text{2}}\)la so nguyen
\(\text{Tìm }x,\text{ }\text{biết:}\)
\(1\text{)}\text{ }2.\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{2}\right)=\frac{1}{2}x\)
\(2\text{) }-3\left(x-\frac{1}{4}\right)-\frac{1}{3}\left(x+\frac{1}{2}\right)=x\)
\(\text{3) }\frac{3}{2}\left(x-\frac{5}{3}\right)-\frac{4}{5}=x+1\)
\(\text{4) }\frac{1}{6}\left(2.x-3\right)=\frac{1}{2}\left(-x+\frac{1}{4}-\frac{2}{3}\right)\)
\(\text{5) }-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2.x-1\right)\)
\(\frac{11}{40}+\frac{4}{5}-\text{|}\frac{3}{4}-1\frac{5}{12}\text{|}=\frac{3}{20}-x\)
\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}.\text{{}\text{[}5.\left(8-\frac{13}{15}\right)\text{]}-1\frac{1}{3}\)
Các bạn giải giúp mình
1.So sánh A và B:
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
2.tính tổng
A=1+2+22+...+2100
B=3-32+33-3+...+399-3100
C1+52+54+56+...+5200
D=7-74+77+...+7301
E=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
F=\(0-\frac{4}{\text{5}}+\frac{4}{\text{5}^2}-\frac{4}{\text{5}^3}+...+\frac{4}{\text{5}^{200}}\)
\(\text{Cho }P=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{399}{400}\text{ Chứng minh }P< \frac{1}{20}\)
chứng tỏ rằng: A= \(\frac{\text{1}}{2\text{1}}\) +\(\frac{2}{3\text{1}}\) +...+ \(\frac{20\text{1}3}{20\text{1}4}\) < 1
