\(1^2-2^2+3^2-4^2+..+2017^2-2018^2+2019^2\)
\(=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2019^2-2018^2\right)\)
\(=1+\left(3+2\right)\left(3-2\right)+\left(5+4\right)\left(5-4\right)+...+\left(2019+2018\right)\left(2019-2018\right)\)
\(=1+2+3+4+5+...+2018+2019\)
\(=\left(1+2019\right).2019\)
\(=4078380\)
https://hoc24.vn/hoi-dap/question/954739.html
Làm rồi mà :D
\(Thugọn:1^2-2^2+3^2-4^2+...+2017^2-2018^2+2019\)
Tính: \(1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...+2016^2+2017^2+2018^2-2019^2-2020^2\)
Tính
\(A=1^2+2^2+3^2-4^2-5^2+6^2+7^2+8^2-9^2-10^2+...+2016^2+2017^2+2018^2-2019^2-2020^2\)
Rút gọn biểu thức S = \(\frac{2019}{2\sqrt{1}+1\sqrt{2}}+\frac{2019}{3\sqrt{2}+2\sqrt{3}}+\frac{2019}{4\sqrt{3}+3\sqrt{4}}+...+\frac{2019}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Mk chỉ cần kết quả thôi , cảm ơn nhiều ạ
Giải phương trình:
\(\sqrt[3]{3x^2-2x+2017}-\sqrt[3]{3x^2-8x+2018}-\sqrt[3]{6x-2019}=\sqrt[3]{2018}\)
Tính tổng
A=12+22+32-42-52+62+72+82-92-102+...+20162+20172+20182-20192-20202
Chứng minh \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{2018\sqrt{2017}}+\dfrac{1}{2019\sqrt{2018}}\)
Cho a,b,c thỏa mãn\(\frac{2}{\left(x^2+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\) .
Tính M=\(\frac{a^{2017}+b^{2018}+c^{2918}}{a^{2017}b^{2018}c^{2019}}\)
Tính \(C=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
