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5 tháng 4 2020

\(1^2-2^2+3^2-4^2+..+2017^2-2018^2+2019^2\)

\(=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2019^2-2018^2\right)\)

\(=1+\left(3+2\right)\left(3-2\right)+\left(5+4\right)\left(5-4\right)+...+\left(2019+2018\right)\left(2019-2018\right)\)

\(=1+2+3+4+5+...+2018+2019\)

\(=\left(1+2019\right).2019\)

\(=4078380\)

5 tháng 4 2020

https://hoc24.vn/hoi-dap/question/954739.html

Làm rồi mà :D

5 tháng 4 2020

=(1-2)(1+2)+(3-4)(3+4)+...+(2017-2018)(2017+2018)+2019

=-(1+2+3+...+2018)+2019

=\(-\frac{2019.2018}{2}+2019\)

\(=-2019.1009+2019\)

=-1008.2019

545002262

31 tháng 7 2018

a/ Ta có:

\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)

\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)

31 tháng 7 2018

a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)

12 tháng 9 2018

\(x=1-\sqrt[2]{2}+\sqrt[2]{4}\)

\(\Leftrightarrow x\left(\sqrt[3]{2}+1\right)=\left(1-\sqrt[2]{2}+\sqrt[2]{4}\right)\left(\sqrt[3]{2}+1\right)=3\)

\(\Leftrightarrow\sqrt[3]{2}x=3-x\)

\(\Leftrightarrow2x^3=27-27x+9x^2-x^3\)

\(\Leftrightarrow x^3-3x^2+9x-9=0\)

Giờ tự rap xô vô nhe

a: Đặt a=2017

\(A=\sqrt{1+\left(\dfrac{1}{a}+\dfrac{1}{a+2}\right)^2}\)

\(=\sqrt{1+\left(\dfrac{2a+2}{a\left(a+2\right)}\right)^2}\)

\(=\sqrt{1+\dfrac{4a^2+8a+4}{a^2\cdot\left(a+2\right)^2}}=\sqrt{\dfrac{\left(a^2+a\right)^2+4a^2+8a+4}{a^2\left(a+2\right)^2}}\)

\(=\sqrt{\dfrac{\left(a^2+a\right)^2+4\left(a+1\right)^2}{a^2\left(a+2\right)^2}}\)

\(=\dfrac{\sqrt{\left(a^2+a\right)^2+4\left(a+1\right)^2}}{a\left(a+2\right)}\)

\(=\dfrac{\sqrt{\left(2017^2+2017\right)^2+4\cdot2018^2}}{2017\cdot2019}\)

b: Đặt 2017=a

\(B=\sqrt{a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2}\)

\(=\sqrt{2a^2+2a+1+\left(a^2+a\right)^2}\)

\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)

\(=2017^2+2017+1=4070307\)