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\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
= \(\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+1-\dfrac{1}{4}+...+1-\dfrac{1}{100}}\)
= \(\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
=\(\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)= 2
c) gọi biểu thức là S = 2 + 2\(^2+2^3+.....+2^{50}\)
2S=2\(^2+2^3+2^4+......+2^{50}+2^{51}\)
\(2S-S=S=2^{51}-2\)
b) \(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{10}}\)
= \(2+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^9}\)
2S-S=S=(\(2+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^9}\))-( \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\))
bạn tự tìm S nhé
mink làm được như thế đó, phần a mink không muốn nhấn mỏi tay bạn ạ, đừng nghĩ mink ko biết làm nha
3:
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
A= 1/3 + 1/3^2 + ... + 1/3^8
3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)
3A=1+ 1/3 + 1/3^2+ ... +1/3^7
=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)
=> 2A= 1 - 1/ 3^8
2A= 6560/6561
A= 6560/6561 : 2
A= 3280/6561
\(\dfrac{1}{A}=\dfrac{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\)
\(\dfrac{1}{A}=1-\dfrac{\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\)
\(\dfrac{1}{A}=1-\dfrac{\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}{\dfrac{3.4+2.4-2.3}{2.3.4}}\)
\(\dfrac{1}{A}=\dfrac{1}{3.4+2.4-2.3}\)
\(\dfrac{1}{A}=1-\dfrac{1}{14}\) \(=\dfrac{13}{14}\)
⇒ \(A=\dfrac{14}{13}\)
Cách 2:
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) ( 1 )
Có: \(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)\(=\dfrac{12}{24}+\dfrac{8}{24}-\dfrac{6}{24}=\dfrac{14}{24}\)
Thay \(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\) \(=\dfrac{14}{24}\) vào ( 1 ), ta có:
\(\dfrac{\dfrac{14}{24}}{\dfrac{14}{24}-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) \(=\dfrac{\dfrac{14}{24}}{\dfrac{14}{24}-\dfrac{1}{24}}\) \(=\dfrac{\dfrac{14}{24}}{\dfrac{13}{24}}\) \(=\dfrac{14}{24}:\dfrac{13}{24}=\dfrac{14.24}{13.24}=\dfrac{14}{13}\)
Vậy \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}}\) \(=\dfrac{14}{13}\).
5: \(=\left(1+2+3+4-3-2-1\right)+\left(-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)\)
\(=4+\left(-1-1-1\right)=4-3=1\)
6: \(=\dfrac{25-49}{35}-\left[\dfrac{1}{2}+\dfrac{2}{7}+\dfrac{1}{10}\right]\)
\(=\dfrac{-24}{35}-\dfrac{35+20+7}{70}\)
\(=\dfrac{-24}{35}-\dfrac{62}{70}=\dfrac{-48-62}{70}=-\dfrac{110}{70}=-\dfrac{11}{7}\)
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
Tính P = 11+2+11+2+3+11+2+3+4+...+11+2+3+4+...+2021
Chúc bạn học tốt nhé
P=1+1/3+1/6+1/10+…..+1/2021×2022÷2
P/2=1/2+1/6+1/12+1/20+…..+1/2021×2022
P/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022
P/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022
P=2021/1011
Chúc bn học tốt