\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)

\(=\sqrt{\frac{5-2\sqrt{5}+1}{2}}+\sqrt{\frac{5+2\sqrt{5}+1}{2}}\)

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{5}-1}{\sqrt{2}}+\frac{\sqrt{5}+1}{\sqrt{2}}\)

\(=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

\(0< x,y,z< 4\)\(\Rightarrow\)\(\hept{\begin{cases}x\left(x-4\right)< 0\\y\left(y-4\right)< 0\\z\left(z-4\right)< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2< 4x\\y^2< 4y\\z^2< 4z\end{cases}\Leftrightarrow}\hept{\begin{cases}x^3>\frac{x^4}{4}\\y^3>\frac{y^4}{4}\\z^3>\frac{z^4}{4}\end{cases}}}\)

\(\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}>\sqrt[4]{\frac{x^4}{4}}+\sqrt[4]{\frac{y^4}{4}}+\sqrt[4]{\frac{z^4}{4}}=\frac{x+y+z}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)

MN ƠI GIÚP MK NHA

Ta có:\(x\left(x+1\right)=k\left(k+2\right)\)

\(\Rightarrow x^2+x=k^2+2k\)

\(\Rightarrow x^2+x+1=k^2+2k+1\)

\(\Rightarrow x^2+x+1=\left(k+1\right)^2\)

Lại có:

\(x^2+x+1< x^2+2x+1=\left(x+1\right)^2\left(1\right)\) vì \(x\in Z^+\)

\(x^2+x>x^2\left(2\right)\)vì \(x\in Z^+\)

Từ \(\left(1\right);\left(2\right)\Rightarrow x^2< x^2+x+1< \left(x+1\right)^2\)

\(\Rightarrow x^2< \left(k+1\right)^2< \left(x+1\right)^2\)

Do \(\left(k+1\right)^2\) là số chính phương bị kẹp giữa 2 số chính phương liên tiếp nên không tồn tại k;x thỏa mãn đề bài

Đề bài => \(c\ge0\)

Đặt \(t=x+\frac{a+b}{2}\)

=> \(\left(t+\frac{a-b}{2}\right)^4+\left(t-\frac{a-b}{2}\right)^4=c\)

<=> \(2t^4+\frac{6t^2\left(a-b\right)^2}{4}.2+\frac{\left(a-b\right)^4}{8}=c\)

<=> \(2t^4+3t^2\left(a-b\right)^2+\frac{\left(a-b\right)^4}{8}-c=0\left(1\right)\)

Ta có \(\Delta=9\left(a-b\right)^4-\left(a-b\right)^4+8c=8\left(a-b\right)^4+8c\ge0\)

=> \(\left(a-b\right)^4+c\ge0\)luôn đúng \(\forall c\ge0\)

Để PT ban đầu có nghiệm 

thì Pt (1) có ít nhất 1 nghiệm dương

=> \(\frac{-3\left(a-b\right)^2+\sqrt{\left(a-b\right)^4+c}}{4}\ge0\)

=> \(c\ge8\left(a-b\right)^4\)

Vậy Pt ban đầu có nghiệm khi \(c\ge8\left(a-b\right)^4\ge0\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right).\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)|\sqrt{5}-\sqrt{3}|\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)

\(=2.[\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)]\)

\(=2.\left(4^2-\sqrt{15}^2\right)\)

\(=2.1=2\)

a) tìm x ể e xác định rồi rút gọn E

b) tìm x để E = \(\frac{-1}{2}\)

c) Tìm GTNN của E