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Áp dụng bất đẳng thức Min.cop.xki 

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

Dấu "=" xảy ra khi \(\frac{a}{c}=\frac{b}{d}\) (Chứng minh bằng biến đổi tương đương)

Áp dụng:

\(S=\sqrt{a^2+\frac{1}{b+c}}+\sqrt{b^2+\frac{1}{c+a}}+\sqrt{c^2+\frac{1}{a+b}}\ge\sqrt{\left(a+b\right)^2+\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\right)^2}+\sqrt{c^2+\frac{1}{a+b}}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\right)^2}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)^2}\)

Theo Bunhiacopxki: \(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)\)

\(\Rightarrow\left(a+b+c\right)^2+\frac{81}{\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2}\ge\left(a+b+c\right)^2+\frac{81}{6\left(a+b+c\right)}\)

\(=\frac{\left(a+b+c\right)^2}{32}+\frac{81}{12\left(a+b+c\right)}+\frac{81}{12\left(a+b+c\right)}+\frac{31}{32}\left(a+b+c\right)^2\)

\(\ge3\sqrt[3]{\frac{\left(a+b+c\right)^2}{32}.\frac{81}{12\left(a+b+c\right)}.\frac{81}{12\left(a+b+c\right)}}+\frac{31}{32}.6^2\)

\(=\frac{153}{4}=\left(\frac{3\sqrt{17}}{2}\right)^2\)

\(\Rightarrow S\ge\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=2\).

????????????

Biểu diễn \(P=\left(1998n^2+1998n\right)+\left(n^2-n+30\right)..\)

Vì \(\left(1998n^2+1998n\right)⋮6n;....P⋮6n\)\(\Leftrightarrow\left(n^2-n+30\right)⋮6n\)

Xét 2 trường hợp 

. Nếu \(n>0:\)

Ta có \(\left(n^2-n\right)⋮n\)\(\Rightarrow30⋮n\)(1)

Lại có \(30⋮6\Rightarrow\left(n^2-n\right)⋮6\)

Mà \(n^2-n=n\left(n-1\right)⋮2\Rightarrow n \left(n-1\right)⋮3\)

\(\Rightarrow n=3k\)hoặc \(n=3k+1\)

Vậy \(P⋮6n\Leftrightarrow n=3k\)hoặc \(n=3k+1\)và \(30⋮n\)(theo (1) )

\(\Rightarrow n\in\left\{1;3;10;30\right\}.\)

. Nếu \(n< 0\)Đặt \(n=-m\)với \(m>0\)

Làm tương tự, ta có \(m\in\left\{2;5;6;15\right\}\Rightarrow n\in\left\{-2;-5;-6;-15\right\}.\)

1+6/3²+4²=7/25

a/

Ta có: \(\frac{1}{2}\left[\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\right]\ge0\)

\(\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Dấu "=" xảy ra khi a = b = c.

b/ Áp dụng câu a.

\(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge\sqrt{\frac{bc}{a}.\frac{ca}{b}}+\sqrt{\frac{ca}{b}.\frac{ab}{c}}+\sqrt{\frac{ab}{c}.\frac{bc}{a}}=a+b+c\)

Dấu "=" xảy ra khi a = b = c.

Cách làm tắt:

a/ \(\sqrt{x}\left(\sqrt{x}+1\right)=6\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow\sqrt{x}=2\text{ (do }\sqrt{x}+3>0\text{)}\)

\(\Leftrightarrow x=4\)

b/ \(x-4-2\sqrt{x-1}=0\Leftrightarrow\left(x-1\right)-2\sqrt{x-1}-3=0\)\(\Leftrightarrow\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}-3\right)=0\)

\(\Leftrightarrow\sqrt{x-1}=3\text{ (do }\sqrt{x-1}+1>0\text{)}\)

\(\Leftrightarrow x=3^2+1=10\)