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Ta có
\(\hept{\begin{cases}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\left(1\right)\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\left(2\right)\end{cases}}\)
\(\Rightarrow\frac{1}{\sqrt{2008}+\sqrt{2005}}+\frac{1}{\sqrt{2010}+\sqrt{2007}}>\frac{2}{\sqrt{2015}+\sqrt{2009}}\)
\(\Leftrightarrow\frac{\sqrt{2008}-\sqrt{2005}}{3}+\frac{\sqrt{2010}-\sqrt{2007}}{3}>\frac{\sqrt{2015}-\sqrt{2009}}{3}\)
\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)
a. Ta có \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\to\frac{1}{\sqrt{2016}+\sqrt{2015}}<\frac{1}{\sqrt{2015}+\sqrt{2014}}\). Nhân liên hợp từng phân thức, ta có
\(\frac{\sqrt{2016}-\sqrt{2015}}{\left(\sqrt{2016}+\sqrt{2015}\right)\left(\sqrt{2016}-\sqrt{2015}\right)}<\frac{\sqrt{2015}-\sqrt{2014}}{\left(\sqrt{2015}+\sqrt{2014}\right)\left(\sqrt{2015}-\sqrt{2014}\right)}\)
\(\Leftrightarrow\sqrt{2016}-\sqrt{2015}<\sqrt{2015}-\sqrt{2014}\Leftrightarrow\sqrt{2016}+\sqrt{2014}<2\sqrt{2015}.\)
b. Tiếp tục thực hiện các biến đổi liên hợp, ta có
\(\sqrt{2008}-\sqrt{2005}+\sqrt{2009}-\sqrt{2007}=\frac{3}{\sqrt{2008}+\sqrt{2005}}+\frac{2}{\sqrt{2009}+\sqrt{2007}}\)
\(>\frac{3}{\sqrt{2015}+\sqrt{2010}}+\frac{2}{\sqrt{2015}+\sqrt{2010}}=\frac{5}{\sqrt{2015}+\sqrt{2010}}=\sqrt{2015}-\sqrt{2010}\)
Suy ra \(\sqrt{2008}-\sqrt{2005}+\sqrt{2009}-\sqrt{2007}>\sqrt{2015}-\sqrt{2010}\to\)
\(\to\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}.\) (ĐPCM).
\(C=\sqrt[3]{2011}-\sqrt[3]{2010}=\frac{2011-2010}{\left(\sqrt[3]{2011^2}+\sqrt[3]{2011}\sqrt[3]{2010}+\sqrt[3]{2010^2}\right)}=\frac{1}{\left(\sqrt[3]{2011^2}+\sqrt[3]{2011}\sqrt[3]{2010}+\sqrt[3]{2010^2}\right)}\)
\(B=\sqrt[3]{2010}-\sqrt[3]{2009}=\frac{2010-2009}{\left(\sqrt[3]{2010^2}+\sqrt[3]{2010}\sqrt[3]{2009}+\sqrt[3]{2009^2}\right)}=\frac{1}{\left(\sqrt[3]{2010^2}+\sqrt[3]{2010}\sqrt[3]{2009}+\sqrt[3]{2009^2}\right)}\)Vì \(\left(\sqrt[3]{2011^2}+\sqrt[3]{2011}\sqrt[3]{2010}+\sqrt[3]{2010^2}\right)>\left(\sqrt[3]{2010^2}+\sqrt[3]{2010}\sqrt[3]{2009}+\sqrt[3]{2009^2}\right)\)
\(B< C\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(A-B=\sqrt{2009}-\sqrt{2007}+\sqrt{2010}-\sqrt{2008}+\sqrt{2011}-\sqrt{2015}\)
\(=\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}\)
Ta có \(\left\{{}\begin{matrix}\sqrt{2009}+\sqrt{2007}< \sqrt{2011}+\sqrt{2015}\\\sqrt{2010}+\sqrt{2008}< \sqrt{2011}+\sqrt{2015}\end{matrix}\right.\)
\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}>\frac{2}{\sqrt{2011}+\sqrt{2015}}+\frac{2}{\sqrt{2011}+\sqrt{2015}}=\frac{4}{\sqrt{2011}+\sqrt{2015}}\)
\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}>0\)
\(\Rightarrow A-B>0\Rightarrow A>B\)