\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

= \(\sqrt{7}-1-\left(1+\sqrt{7}\right)\)

\(=\sqrt{7}-1-\sqrt{7}-1\)

= -2

1.  \(2\sqrt{5}-5\sqrt{20}+\sqrt{80}\)

= \(2\sqrt{5}-5.2\sqrt{5}+4\sqrt{5}\)

= \(2\sqrt{5}-10\sqrt{5}+4\sqrt{5}\)

= \(-4\sqrt{5}\)

2. B = \(\frac{1}{\sqrt{5}-2}-\sqrt{6-2\sqrt{5}}\)

       = \(\frac{1}{\sqrt{5}-2}-\sqrt{1^2-2\sqrt{5}+\sqrt{15}^2}\)

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

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

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

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

         = \(\frac{-6+3\sqrt{5}}{\sqrt{5}-2}=\frac{3\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=3\)

\(x=2\sqrt{2}+3=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)

Thay vào A ta được:

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

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

\(\frac{2\sqrt{2}-2+4-2\sqrt{2}}{2-1}=2\)

Ta có \(\frac{a^3}{b^2+1}=\frac{a^3}{b^2+ab}=\frac{a^3}{b\left(a+b\right)}\)

Áp dụng BĐT cosi

\(\frac{a^3}{b\left(a+b\right)}+\frac{b}{2}+\frac{a+b}{4}\ge\frac{3}{2}a\)

TT \(\frac{b^3}{a\left(a+b\right)}+\frac{a}{2}+\frac{a+b}{4}\ge\frac{3}{2}b\)

=> \(VT\ge\frac{1}{2}\left(a+b\right)\ge\sqrt{ab}=1\)

Dấu bằng xảy ra khi a=b=1

Sửa đề \(A=\frac{x^2}{x+y^2}+\frac{y^2}{y+z^2}+\frac{z^2}{z+x^2}\)

Ta có \(\frac{x^2}{x+y^2}=\frac{x^2+xy^2-xy^2}{x+y^2}=x-\frac{xy^2}{x+y^2}\)

Mà \(x+y^2\ge2y\sqrt{x}\)

=> \(\frac{x^2}{x+y^2}\ge x-\frac{\sqrt{x}}{2}\ge\frac{3x}{4}-\frac{1}{4}\)

=> \(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)

Min A=3/2 khi x=y=z=1

Ta có: \(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2-ab+b^2\ge ab\)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)(Vì a , b > 0)

\(\Rightarrow a^3+b^3\ge a^2b+ab^2\)

\(\Rightarrow a^3\ge b^3-a^2b+ab^2\)

\(\Rightarrow3a^3\ge2a^3-b^3+a^2b+ab^2\)

\(\Rightarrow3a^3\ge a^3-b^3+a^3+a^2b+ab^2\)

\(\Rightarrow3a^3\ge\left(a-b\right)\left(a^2+ab+b^2\right).a\left(a^2+ab+b^2\right)\)

\(\Rightarrow3a^3\ge\left(a^2+ab+b^2\right)\left(2a-b\right)\)

\(\Rightarrow\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\)(1)

Chứng minh tương tự ta có:

\(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b-c}{3}\)(2)

\(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c-a}{3}\)(3)

Cộng vế với vế của (1) , (2) , (3)\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{2a-b+2b-c+2c-a}{3}=\frac{a+b+c}{3}\left(đpcm\right)\)

VÌ A là số nguyên , x nguyên

=> \(\sqrt{x-4}\)là số nguyên

\(A=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x-8+8}{\sqrt{x-4}}=2\sqrt{x-4}+\frac{8}{\sqrt{x-4}}\)là số nguyên

=> \(\frac{8}{\sqrt{x-4}}\)là số nguyên

=> \(\sqrt{x-4}\in\left\{1;2;4;8\right\}\)

=> \(x\in\left\{5;8;20;68\right\}\)

Vậy \(x\in\left\{5;8;20;68\right\}\)