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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: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
a) \(2-2\sqrt{3}\) và \(4-\sqrt{15}\)
Giả sử : \(2-2\sqrt{3}\ge4-\sqrt{15}\)
⇔ \(\sqrt{15}-2\sqrt{3}\ge2\)
⇔ \(\left(\sqrt{15}-2\sqrt{3}\right)^2\ge2^2\)
⇔ 15 - \(12\sqrt{5}+12\) ≥ 4
⇔ 27 -4 ≥ \(12\sqrt{5}\)
⇔ 23 ≥ \(12\sqrt{5}\)
⇔ \(23^2\) ≥ \(\left(12\sqrt{5}\right)^2\)
⇔ 529 ≥ 720 (sai)
Vậy 2 - \(2\sqrt{3}< 4-\sqrt{15}\)
b) \(\sqrt{11}+2\) và \(3+\sqrt{3}\)
Giả sử : \(\sqrt{11}+2\le3+\sqrt{3}\)
⇔ \(\sqrt{11}-\sqrt{3}\le1\)
⇔ \(\left(\sqrt{11}-\sqrt{3}\right)^2\le1\)
⇔ 14 - \(2\sqrt{33}\) ≤ 1
⇔ 13 ≤ \(2\sqrt{33}\)
⇔ \(13^2\le\left(2\sqrt{33}\right)^2\)
⇔ 169 ≤ 132 (sai)
Vậy \(\sqrt{11}+2\ge3+\sqrt{3}\)
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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= \(a^{2017}\left(a^2-8a+11\right)+b^{2017}\left(b^2-8b+11\right)=\)\(a^{2017}\left(a^2-8a+16-5\right)+b^{2017}\left(b^2-8b+16-5\right)=\)\(a^{2017}\left(\left(a-4\right)^2-\sqrt{5^2}\right)+b^{2017}\left(\left(b-4\right)^2-\sqrt{5^2}\right)\)=\(a^{2017}\left(a-4-\sqrt{5}\right)\left(a-4+\sqrt{5}\right)+b^{2017}\left(b-4-\sqrt{5}\right)\left(b-4+\sqrt{5}\right)\)= 0+0= 0
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.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
\(x=1-\sqrt[2]{2}+\sqrt[2]{4}\)
\(\Leftrightarrow x\left(\sqrt[3]{2}+1\right)=\left(1-\sqrt[2]{2}+\sqrt[2]{4}\right)\left(\sqrt[3]{2}+1\right)=3\)
\(\Leftrightarrow\sqrt[3]{2}x=3-x\)
\(\Leftrightarrow2x^3=27-27x+9x^2-x^3\)
\(\Leftrightarrow x^3-3x^2+9x-9=0\)
Giờ tự rap xô vô nhe
A > B