Đặt \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)

Không mất tính tổng quát giả sử \(a\ge b\ge c\ge d\)=>\(a^2\ge b^2\ge c^2\ge d^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)

=>\(A\le\frac{4}{d^2}\)=>\(d^2\le4\)=>\(d\in\text{ }\text{{}\pm1,\pm2\text{ }\)

Xét \(d=\pm1\)=> vô lí

Xét d=\(\pm\)2=> a=b=c=d=\(\pm\)2

=> M=ab+cd=4+4=8

Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+a+b+c=2+2018\)

\(\Leftrightarrow\frac{a+ab+bc}{b+c}+\frac{b+bc+ab}{c+a}+\frac{c+ac+bc}{a+b}=2020\)

\(\Leftrightarrow a\left(\frac{1+b+c}{b+c}\right)+b\left(\frac{1+a+c}{a+c}\right)+c\left(\frac{1+a+b}{a+b}\right)=2020\left(1\right)\)

Vì \(a+b+c=2018\Rightarrow\hept{\begin{cases}a+b=2018-c\\b+c=2018-a\\c+a=2018-b\end{cases}\left(2\right)}\)

Thay (2) vào (1) ta được: 

\(a\left(\frac{2019-a}{b+c}\right)+b\left(\frac{2019-b}{a+c}\right)+c\left(\frac{2019-c}{a+b}\right)=2020\)

\(\Leftrightarrow\frac{2019a-a^2}{b+c}+\frac{2019b-b^2}{a+c}+\frac{2019c-c^2}{a+b}=2020\)

\(\Leftrightarrow\frac{2019a}{b+c}-\frac{a^2}{b+c}+\frac{2019b}{a+c}-\frac{b^2}{a+c}+\frac{2019c}{a+b}-\frac{c^2}{a+b}=2020\)

\(\Leftrightarrow2019\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)

\(\Leftrightarrow4038-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)( vì \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\))

\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=2018\)

\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1=2019\)

b) trước hết ta cần chứng minh nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

ta có x+y+z=0==> x=-(y+z) 

             <=> \(x^3=-\left(y^3+z^3+3yz\left(y+z\right)\right)\)

           <=> \(x^3+y^3+z^3=-3yz\left(y+z\right)\)

      <=> \(x^3+y^3+z^3=3xyz\)( cì y+z=-x)

 áp dụng vào bài ta có \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

 do đó M=\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

neu de bai bai 1 la tinh x+y thi mik lam cho

đăng từng này thì ai làm cho 

We have \(P=\frac{x^4+2x^2+2}{x^2+1}\)

\(\Rightarrow P=\frac{x^4+2x^2+1+1}{x^2+1}\)

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

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

\(\ge2\sqrt{\frac{x^2+1}{x^2+1}}=2\)

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(P_{min}=2\Leftrightarrow x=0\)