\(\sqrt{2017}+\sqrt{2019}và2\sqrt{2018}\)
So sánh:
a) x=\(\sqrt{2017}-\sqrt{2018}\)và y=\(\sqrt{2016}-\sqrt{2017}\)
b) x=\(\sqrt{2019}+\sqrt{2017}\)và y=\(2\sqrt{2018}\)
a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
Giải phương trình:
\(\sqrt[3]{3x^2-2x+2017}-\sqrt[3]{3x^2-8x+2018}-\sqrt[3]{6x-2019}=\sqrt[3]{2018}\)
Rút gọn biểu thức: A= \(\frac{\sqrt{x-2017-2\sqrt{x-2018}}}{\sqrt{x-2018}-1}\)Với x > 2019
Bài 1: Rút gọn biểu thức sau:
a. \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}\)
b. \(B=\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2018\sqrt{2017}+2017\sqrt{2018}}\)
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
b/ \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2017}}-\dfrac{1}{\sqrt{2018}}=1-\dfrac{1}{\sqrt{2018}}\)
So sanh: x=\(\sqrt{2019}\) va y=\(2\sqrt{2018}-\sqrt{2017}\)
So sanh: x=\(\sqrt{2019}\)va y=\(2\sqrt{2018}-\sqrt{2017}\)
Giả sử \(\sqrt{2009}\ge2\sqrt{2008}-\sqrt{2007}\)
\(\Leftrightarrow\sqrt{2009}-\sqrt{2008}\ge\sqrt{2008}-\sqrt{2007}\)
\(\Leftrightarrow\frac{1}{\sqrt{2009}+\sqrt{2008}}\ge\frac{1}{\sqrt{2008}+\sqrt{2007}}\) (sai)
Vậy \(\sqrt{2009}< 2\sqrt{2008}-\sqrt{2007}\)
Không dùng máy tính so sánh \(\sqrt{2019}-\sqrt{2018}\) và\(\sqrt{2018}-\sqrt{2017}\)
Lời giải:
\(\sqrt{2019}-\sqrt{2018}=\frac{2019-2018}{\sqrt{2019}+\sqrt{2018}}=\frac{1}{\sqrt{2019}+\sqrt{2018}}\)
\(\sqrt{2018}-\sqrt{2017}=\frac{2018-2017}{\sqrt{2018}+\sqrt{2017}}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2019}+\sqrt{2018}>\sqrt{2018}+\sqrt{2017}\Rightarrow \frac{1}{\sqrt{2019}+\sqrt{2018}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
\(\Rightarrow \sqrt{2019}-\sqrt{2018}< \sqrt{2018}-\sqrt{2017}\)
Giải phương trình:
x=\(\frac{1}{\sqrt{2019}-\sqrt{2018}}\)và y=\(\frac{1}{\sqrt{2018}-\sqrt{2017}}\)
b,So sánh
a, x=\(\frac{1\left(\sqrt{2019}+\sqrt{2018}\right)}{2019-2018}\) và y=\(\frac{1\left(\sqrt{2018}+\sqrt{2017}\right)}{2018-2017}\) (Trục căn thức ở mẫu)
\(\Leftrightarrow\) x=\(\sqrt{2019}+\sqrt{2018}\) và y=\(\sqrt{2018}+\sqrt{2017}\)
b, Ta có : x - y = (\(\sqrt{2019}+\sqrt{2018}\) ) - ( \(\sqrt{2018}+\sqrt{2017}\) )
= \(\sqrt{2019}-\sqrt{2017}\) > 0
\(\Rightarrow\) x - y > 0 \(\Leftrightarrow\) x > y
\(\frac{1+2017\sqrt{2018}\:-2018\sqrt{2017}}{\sqrt{2017\:\:}+\sqrt{2018}+\sqrt{2017}\cdot\sqrt{2018}}=\sqrt{2017.2018\:}\)