Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right)\left(2018+1\right)}=4036+2\sqrt{2018^2-1}\)
Mặt khác: \(\left(2\sqrt{2018}\right)^2=4.2018=2.2018+2\sqrt{2018^2}=4036+2\sqrt{2018^2}\)
Vì \(4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}\)
Vậy ......
So sánh:
a) x=\(\sqrt{2017}-\sqrt{2018}\)và y=\(\sqrt{2016}-\sqrt{2017}\)
b) x=\(\sqrt{2019}+\sqrt{2017}\)và y=\(2\sqrt{2018}\)
Rút gọn biểu thức: A= \(\frac{\sqrt{x-2017-2\sqrt{x-2018}}}{\sqrt{x-2018}-1}\)Với x > 2019
So sanh: x=\(\sqrt{2019}\)va y=\(2\sqrt{2018}-\sqrt{2017}\)
\(\frac{1+2017\sqrt{2018}\:-2018\sqrt{2017}}{\sqrt{2017\:\:}+\sqrt{2018}+\sqrt{2017}\cdot\sqrt{2018}}=\sqrt{2017.2018\:}\)
rút gon:\(\frac{1+2019\sqrt{2018}-2018\sqrt{2019}}{\sqrt{2018}+\sqrt{2019}+\sqrt{2018.2019}}\)
Cho \(x=1-\sqrt[3]{2}+\sqrt[3]{4}\)Tính \(B=x^{2019}-3x^{2018}+9x^{2017}-9x^{2016}+2019\)
cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
Tìm giá trị nhỏ nhất của biểu thức
\(C=\sqrt{\left(x+2017\right)^2}+\sqrt{\left(x+2018\right)^2}+\sqrt{\left(x+2019\right)^2}\)
So sánh
\(\sqrt{2017}+\sqrt{2019}\)và \(2\sqrt{2018}\)
GIúp mình với sắp kiểm tra rùi
