a. Đề là \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\) ?

\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)

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

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

\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)

\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)

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

b. Thay a = 3b vào Q, ta được : \(Q=\sqrt{\frac{3b-b}{3b+b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}\)

đk: \(x\ge\sqrt{6}\)

Đặt \(\sqrt{x^2-6}=a\Rightarrow x^2-5=a^2-1\)

\(Pt\Leftrightarrow a^2-1+a=7\)

\(\Leftrightarrow a^2+a-8=0\)

\(\Leftrightarrow\left(a-\frac{-1+\sqrt{33}}{2}\right)\left(a-\frac{-1-\sqrt{33}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{-1+\sqrt{33}}{2}\\a=\frac{-1-\sqrt{33}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a^2=\frac{17-\sqrt{33}}{2}\\a^2=\frac{17+\sqrt{33}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{29-\sqrt{33}}{2}\\x^2=\frac{29+\sqrt{33}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{29-\sqrt{33}}{2}}\\x=\pm\sqrt{\frac{29+\sqrt{33}}{2}}\end{cases}}\)

Bài 2: Ta có 2 đẳng thức ngược chiều: \(\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}\ge8;\frac{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^3}\le8\)

Áp dụng BĐT AM-GM ta có:

\(\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+\frac{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^3}\)\(\ge2\sqrt{\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}.\frac{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+c\right)^3}}\)

Suy ra BĐT đã cho là đúng nếu ta chứng minh được

\(27\left(a^2+b^2+c^2\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\left(ab+bc+ca\right)\left(a+b+c\right)^3\left(1\right)\)

Sử dụng đẳng thức \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)và theo AM-GM: \(abc\le\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)ta được \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\left(2\right)\)

Từ (1)và(2) suy ra ta chỉ cần chứng minh \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)đúng=> đpcm

Đẳng thức xảy ra khi và chỉ khi a=b=c

Bài 3:

Ta có 2 BĐT ngược chiều: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2};\sqrt[3]{\frac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\sqrt[3]{\frac{1}{8}}=\frac{1}{2}\)

Bổ đề: \(x^3+y^3+z^3+3xyz\ge xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)\left(1\right)\forall x,y,z\ge0\)

Chứng minh: Không mất tính tổng quát, giả sử \(x\ge y\ge z\). Khi đó:

\(VT\left(1\right)-VP\left(1\right)=x\left(x-y\right)^2+z\left(y-z\right)^2+\left(x-y+z\right)\left(x-y\right)\left(y-z\right)\ge0\)

Áp dụng BĐT AM-GM ta có:

\(\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\ge64\left(abc\right)^2\)\(\Leftrightarrow\frac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\left[\frac{4abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right]^3\)

Suy ra ta chỉ cần chứng minh \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{4abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge2\)

\(\Leftrightarrow a\left(a+b\right)\left(a+c\right)+b\left(b+c\right)\left(b+a\right)+c\left(c+a\right)\left(c+b\right)+4abc\)\(\ge2\left(a+b\right)\left(b+c\right)\left(c+a\right)\)\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)đúng theo bổ đề

Đẳng thức xảy ra khi và chỉ khi a=b=c hoặc a=b,c=0 và các hoán vị

\(pt\Leftrightarrow x^2-6+\sqrt{x^2-6}-6=0\)

\(\Leftrightarrow\left(\sqrt{x^2-6}\right)^2+3\sqrt{x^2-6}-2\sqrt{x^2-6}-6=0\)

\(\Leftrightarrow\left(\sqrt{x^2-6}+3\right)\left(\sqrt{x^2-6}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-6}=-3\left(L\right)\\\sqrt{x^2-6}=2\end{cases}}\)

\(\Rightarrow x^2-6=4\Rightarrow x^2=10\)

\(\Rightarrow x=\sqrt{10}\)

Vơi \(n=2\Rightarrow n^n-n^2+n-1=1\)và \(\left(n-1\right)^2=\left(2-1\right)^2=1\)

\(\Rightarrow n^n-n^2+n-1⋮\left(n-1\right)^2\)

Với n>2 ta có: \(B=\left(n^n-n^2\right)+\left(n-1\right)\)

\(=n^2\left(n^{n-2}-1\right)+\left(n-1\right)\)\(=n^2\left(n-1\right)\left(n^{n-3}+n^{n-4}+...+1\right)+\left(n-1\right)\)

\(=\left(n-1\right)\left(n^{n-1}+n^{n-2}+...+n^2+1\right)\)\(=\left(n-1\right)\text{[}\left(n^{n-1}-1\right)+...+\left(n^2-1\right)+\left(n-1\right)\text{]}\)

\(=\left(n-1\right)^2⋮\left(n-1\right)^2\)(đpcm)