\(Q=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{c+a+b}{abc}=1\)

Ta có

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow3^2=P+2.Q=P+2\Rightarrow P=7\)

\(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(A=\frac{8}{1-x^8}+\frac{8}{1+x^8}\)

\(A=\frac{16}{1-x^{16}}\)

\(A=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)

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

\(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)

\(A=\frac{a^3+b^3+c^3}{2abc}\)

\(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)

\(A=\frac{3abc}{2abc}=\frac{3}{2}\)

a, \(x-1-\frac{x^2-4}{x+1}=x-1-\frac{\left(x-2\right)\left(x+2\right)}{x+1}\)

\(=\frac{\left(x-1\right)\left(x+1\right)}{x+1}-\frac{\left(x-2\right)\left(x+2\right)}{x+1}=\frac{x^2-1-x^2+4}{x+1}=\frac{3}{x+1}\)

c, \(\frac{2x^2+1}{x^3+1}-\frac{x-1}{x^2-x+1}-\frac{1}{x+1}\)

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

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

\(H=2020-x^2-3y^2+2xy-10x+14y\)

\(=2020-x^2+2xy-y^2-2y^2-10x+14y\)

\(=2020-\left(x^2-2xy+y^2\right)-2y^2-10x+14y\)

\(=1995-\left(x-y\right)^2-2.\left(x-y\right).5-25-2y^2+24y\)

\(=1959-\left[\left(x-y\right)^2+2.\left(x-y\right).5+25\right]-2\left(y^2-12y+36\right)\)

\(=1959-\left(x-y+5\right)^2-2\left(y-6\right)^2\)

Vì \(\left(y-6\right)^2\ge0\)

\(\Rightarrow2\left(y-6\right)^2\ge0\)

và \(\left(x-y+5\right)^2\ge0\)

\(\Rightarrow1959-\left(x-y+5\right)^2-2\left(y-6\right)^2\le1959\)

\(\Rightarrow H\le1959\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y+5\right)^2=0\\\left(y-6\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y+5=0\\y-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=6\end{cases}}\)

Vậy GTLN : Max H = 1959 khi x = 1 ; y = 6

H = 2020 - x² - 3y² + 2xy - 10x + 14y

= -x² + 2xy - y² - 10x + 10y - 25 - 2y² + 4y - 2 + 2047

= -(x - y)² - 10(x - y) - 25 - 2(y - 1)² + 2047

= -(x - y + 5)²  - 2(y - 1)² + 2047 \(\le2047\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y+5=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x-y=-5\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-4\\y=1\end{cases}}\)

Vậy Max H = 2047 <=> \(x=-4;y=1\)

N, M là điểm như thế nào vậy bạn?