<  nha

hoc tot

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 \(\sqrt{2009}< 2\sqrt{2008}-\sqrt{2007}\)

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

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

\(\Rightarrow B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2017}}-\dfrac{1}{\sqrt{2018}}=1-\dfrac{1}{\sqrt{2018}}\)

Lời giải:

Câu GPT: bạn xem lại đề bài.

Câu so sánh

Áp dụng hằng đẳng thức: \((a-b)(a+b)=a^2-b^2\Rightarrow a-b=\frac{a^2-b^2}{a+b}\) vào bài toán ta có:

\(\sqrt{2018}-\sqrt{2017}=\frac{2018-2017}{\sqrt{2018}+\sqrt{2017}}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)

\(\sqrt{2019}-\sqrt{2018}=\frac{2019-2018}{\sqrt{2019}+\sqrt{2018}}=\frac{1}{\sqrt{2019}+\sqrt{2018}}\)

Mà dễ thấy \(0< \sqrt{2018}+\sqrt{2017}< \sqrt{2019}+\sqrt{2018}\Rightarrow \frac{1}{\sqrt{2018}+\sqrt{2017}}> \frac{1}{\sqrt{2019}+\sqrt{2018}}\)

\(\Rightarrow \sqrt{2018}-\sqrt{2017}> \sqrt{2019}-\sqrt{2018}\)

Akai Haruma cô ơi em có cách khác câu so sánh mặc dù có lẽ cách này không hay và ngắn gọn như của cô:) (câu gpt thì cách em hệt của cô rồi)

Xét hiệu hai vế: \(\sqrt{2018}-\sqrt{2017}-\sqrt{2019}+\sqrt{2018}\)

\(=2\sqrt{2018}-\left(\sqrt{2019}+\sqrt{2017}\right)\)

\(=2\sqrt{2018}-\frac{2}{\sqrt{2019}-\sqrt{2017}}\)

\(=2\left(\sqrt{2018}-\frac{1}{\sqrt{2019}-\sqrt{2017}}\right)\)

Ta có: \(\sqrt{2018}>1;\sqrt{2019}-\sqrt{2017}>0\Rightarrow\frac{1}{\sqrt{2019}-\sqrt{2017}}< 0\)

Từ đây suy ra \(2\left(\sqrt{2018}-\frac{1}{\sqrt{2019}-\sqrt{2017}}\right)>2\left(1-1\right)=0\)

Suy ra \(\sqrt{2018}-\sqrt{2017}>\sqrt{2019}-\sqrt{2018}\)

a, \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\)-\(\frac{3\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)

=\(\sqrt{2}-3\)

b,X=\(\sqrt{2019}+\sqrt{2018}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2019}+\sqrt{2018}\))

Y=\(\sqrt{2018}+\sqrt{2017}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2018}+\sqrt{2017}\))

So sánh:X & Y<=>X-\(\sqrt{2018}\)&Y-\(\sqrt{2018}\)(Trừ hai vế cho \(\sqrt{2018}\)) <=>\(\sqrt{2019}\)&\(\sqrt{2017}\)

Có:2019>2017

=>\(\sqrt{2019}>\sqrt{2017}\)

=>X>Y

Câu b, mk ko bt có lm đúng ko?

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

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

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

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

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

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

Giờ tự rap xô vô nhe

giai chi tiet giup minh voi 

2018/-2017<-1=-2019/2019

suy ra 2018/-2017<-2019/2019

                            Giải

Ta có: \(\frac{2018}{-2017}=-1+\frac{-1}{2017}\)

\(\Rightarrow\frac{2018}{-2017}>-1\)

Mà \(\frac{-2019}{2019}=1\) nên \(\frac{2018}{-2017}>\frac{-2019}{2019}\)