Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
\(\left(1-1\right)-\left(\frac{1}{2}+\frac{1}{2}\right)+\left(2-2\right)-\left(\frac{2}{3}+\frac{1}{3}\right)+\left(3-3\right)-\left(\frac{3}{4}+\frac{1}{4}\right)-1\)
\(-1-1-1+4=1\)
MIK XIN LỖI BN NHA VÌ ĐÁNH MÁY HƠI LÂU NHA !! CHÚC BN HOK TỐT NHAA
với giải thích hộ mik số trên có chia hết cho 13 ko và có là số chính phương không ạ
Đặt biểu thức trên là A , ta có :
\(A=1+3+3^2+3^3+...+3^{98}\)
\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{99}\)
\(\Leftrightarrow3A-A=\left(3+3^2+3^3+3^4+...+3^{99}\right)-\left(1+3+3^2+3^3+...+3^{98}\right)\)
\(\Leftrightarrow2A=3^{99}-1\)
\(\Leftrightarrow A=\frac{3^{99}-1}{2}\)
\(=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{2021\cdot2022:2}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}=\dfrac{1010}{1011}\)
1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)
\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)
\(\Rightarrow3S-S=3^{2021}-1\)
\(\Rightarrow2S=3^{2021}-1\)
\(\Rightarrow S=\frac{3^{2021}-1}{2}\)
2. \(\left(3x-2\right)^3=64\)
\(\Leftrightarrow\left(3x-2\right)^3=4^3\)
\(\Leftrightarrow3x-2=4\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
[3x-2]^3=64
Ta có:64=4^3
Suy ra:3x-2=4
3x =4+2
3x=6
x=6:3
x=2
Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{A}=1:\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(\Rightarrow\dfrac{1}{A}=2+3+4+5+...+100\)
\(\Rightarrow\dfrac{1}{A}=\left(100+2\right)\cdot99:2\)
\(\Rightarrow\dfrac{1}{A}=5049\)
\(\Rightarrow A=\dfrac{1}{5049}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-.....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{24}{50}=\frac{12}{25}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{12}{25}\)
\(\frac{3}{8}\left(x-\frac{2}{3}\right)-\frac{2}{9}\left(3x-\frac{1}{2}\right)=\frac{-7}{36}\)
\(\Rightarrow\frac{3}{8}x-\frac{1}{4}-\frac{2}{3}x+\frac{1}{9}=\frac{-7}{36}\Rightarrow\frac{-7}{24}x-\frac{5}{36}=\frac{-7}{36}\)
\(\Rightarrow\frac{-7}{24}x=\frac{-1}{3}\Rightarrow x=\frac{8}{7}\)
\(\frac{2}{3}+\frac{1}{2}:\frac{-3}{4}=\frac{2}{3}+\frac{1}{2}.\frac{4}{-3}=\frac{2}{3}+\frac{-2}{3}=0\)
2/3+-2
2/3+-6/3
-4/3