Ta có : \(-x^2+2x-4\)

        \(=-\left(x^2-2x+1\right)-3\)

          \(=-\left(x-1\right)^2-3\)\(\le-3\forall x\)

\(\Rightarrow E=\frac{3}{-x^2+2x-4}\)\(\ge\frac{3}{-3}=-1\forall x\)

\(E=-1\Leftrightarrow-\left(x-1\right)^2=0\)

                 \(\Leftrightarrow x=1\)

Vậy \(MinE=-1\Leftrightarrow x=1\)

Ta có : 

\(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2015+2\sqrt{2015.2017}+2017=8064+2\sqrt{2015.2017}\)

\(\left(2\sqrt{2016}\right)^2=8064\)

Vì \(\left(\sqrt{2015}+\sqrt{2017}\right)^2>\left(2\sqrt{2016}\right)^2\) nên \(\sqrt{2015}+\sqrt{2017}>2\sqrt{2016}\)

Vậy... 

Chúc bạn học tốt ~ 

Cảm ơn bn nhiều nhé :)))

a)

\(A=\cot^2x\left(\cos^2x-1+\sin^2x\right)+\sin^2x\)

\(A=\cot^2x\left(\cos^2x+\sin^2x-1\right)+\sin^2x\)

\(A=\cot^2x\left(1-1\right)+\sin^2x\)

\(A=\cot^2x.0+\sin^2x\)

\(A=\sin^2x\)

b) \(B=\cos^4\alpha-\sin^4\alpha+2\sin^2\alpha+8\)

\(B=\left(cos^2\alpha+sin^2\alpha\right)\left(cos^2\alpha-sin^2\alpha\right)+2\sin^2\alpha+8\)

\(B=cos^2\alpha-sin^2\alpha+2\sin^2\alpha+8\)

\(B=cos^2\alpha+sin^2\alpha+8\)

\(B=1+8\)

\(B=9\)

\(A=\frac{\sqrt{a+b}-\sqrt{a}+\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+\frac{2\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+B\)

\(B=\frac{2\sqrt{b}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{-2\sqrt{ab}}=-\frac{\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}}=-\frac{\sqrt{a}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{a}\)

\(=\frac{\left(\sqrt{a+b}-\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a+b}-\sqrt{a}\right)^2-b}\)

\(=\frac{2a+2b-2\sqrt{a\left(a+b\right)}+2\sqrt{b\left(a+b\right)}-2\sqrt{ab}}{2a-2\sqrt{a\left(a+b\right)}}\)

\(=\frac{a+b-\sqrt{a\left(a+b\right)}+\sqrt{b\left(a+b\right)}-\sqrt{ab}}{a-\sqrt{a\left(a+b\right)}}\)

\(=\frac{\left(a+b-\sqrt{a\left(a+b\right)}+\sqrt{b\left(a+b\right)}-\sqrt{ab}\right)\left(a+\sqrt{a\left(a+b\right)}\right)}{a^2-a\left(a+b\right)}\)

\(=\frac{b\sqrt{a\left(a+b\right)}+\sqrt{ab}\left(a+b\right)-a\sqrt{ab}}{-ab}\)

\(=\frac{-b\sqrt{a\left(a+b\right)}+b\sqrt{ab}}{ab}\)

\(=\frac{\sqrt{ab}-\sqrt{a\left(a+b\right)}}{a}\)

có 2004²=4016016 chia 11 dư 1

=>(2004²)¹⁰⁰²=4.......016 chia 11 dư 1

=> 2004²⁰⁰⁴ chia 11 dư 1

Ta thấy: k thuộc N^* nên \(\sqrt{k+1}>\sqrt{k}\)

\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{2}{\left(2\sqrt{k+1}\right).\left(\sqrt{k+1}.\sqrt{k}\right)}< \frac{2}{\left(\sqrt{k+1}.\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}\)

\(=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}.\sqrt{k}\right)\left(k+1-k\right)}=2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)

\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)(đpcm).