Tính nhanh: \(A=\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}}{500-\frac{500}{501}-\frac{501}{502}-\frac{502}{503}-...-\frac{999}{1000}}\)
\(\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}}{500-\frac{500}{501}-\frac{501}{502}-...-\frac{999}{1000}}=\frac{\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{999}-\frac{1}{1000}\right)}{500-\left(1-\frac{1}{501}\right)-\left(1-\frac{1}{502}\right)-...-\left(1-\frac{1}{1000}\right)}\)
hình như cái mẫu bạn ghi dấu sai thì phải, còn tử thì mình lười làm lắm
tử bạn tính ra 1/2+1/12+...+1/999 000 sau đó phân tích ra là
khó thật
nhớ L-I-K-E nhe tại vì cậu bảo giúp mình, mình cho đúng liền
Tính:
D=\(\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{999}-\frac{1}{1000}}{500-\frac{500}{501}-\frac{501}{502}-.....-\frac{999}{1000}}\)
1-1/2+1/3-1/4+......-1/1000
=(1+1/3+1/5+......+1/999)-(1/2+1/4+.......+1/1000)
=(1+1/2+1/3+1/4+.....+1/1000)-2(1/2+1/4+.......+1/1000)
=(1+1/2+1/3+.........+1/1000)-(1+1/2+.....+1/500)
=1/501 +1/502+1/503+.....+1/1000 ;
mat khác:
500-500/501-501/502-.....-999/1000
=(1-500/501)+(1-501/502)+.....+(1-999/1000)=1/501+1/502+....+1/1000
=>D=1
Chứng minh rằng :
1 - \(\frac{1}{2}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{5}\) - \(\frac{1}{6}\)+ . . . + \(\frac{1}{999}\)-\(\frac{1}{1000}\) = \(\frac{1}{501}\)+ \(\frac{1}{502}\)+\(\frac{1}{503}\) + . . . + \(\frac{1}{1000}\)
Câu 1: Không quy đồng hãy so sánh :
\(\frac{17171718}{19191920}\) và \(\frac{33}{38}\)
Câu 2: Chứng tỏ rằng :
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{999}-\frac{1}{1000}=\frac{1}{501}+\frac{1}{502}+.......+\frac{1}{1000}\)
Câu 3: Tìm x :
\(\frac{3}{4}x-\frac{4}{3}x-\frac{1}{12}x+\frac{7}{48}=0\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+.....+\frac{2016}{501}}{\frac{-1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-.....-\frac{1}{999\cdot1000}}\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)
\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)
\(B=\frac{2016}{-1}=-2016\)
Chứng minh rằng \(\frac{1}{3\sqrt[2]{2}}+\frac{1}{4\sqrt[3]{3}}+\frac{1}{5\sqrt[3]{4}}+.....+\frac{1}{1000\sqrt[3]{999}}< \frac{11}{5}\)
chứng minh rằng \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
Áp dụng tính \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
\(VT=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\left(\frac{2}{ab}-\frac{2}{a\left(a+b\right)}-\frac{2}{b\left(a+b\right)}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2-\frac{2\left(a+b\right)-2b-2a}{ab\left(a+b\right)}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|=VP\)
Áp dụng tính M: \(M=\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
\(M=999.\sqrt{\frac{1}{999^2}+\frac{1}{1^2}+\frac{1}{\left(999+1\right)^2}}+\frac{999}{1000}\)
\(M=999.\left(\frac{1}{1}+\frac{1}{999}-\frac{1}{1000}\right)+\frac{999}{1000}\)
\(M=999+1-\frac{999}{1000}+\frac{999}{1000}=1000\)
Vậy M=1000.
Tính:
\(\left(\frac{1000}{1}+\frac{999}{2}+\frac{998}{3}+\frac{997}{4}+...+\frac{2}{999}+\frac{1}{1000}\right)\)\(:\)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{1000}\right)\)
chứng minh rằng
\(\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{\left(x+y\right)^2}}=\left|\frac{1}{x}+\frac{1}{y}-\frac{1}{x+y}\right|\).áp dụng tính M=\(\sqrt{1+999^2+\frac{999^2}{1000^2}}+\frac{999}{1000}\)
