\(E=5x^2+8xy+5y^2-2x+2y\)

\(=\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)-2\)

\(=4\left(x^2+2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)-2\)

\(=4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2-2\ge-2\) có GTNN là - 2

Dấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\)

Vậy \(E_{min}=-2\) tại \(x=1;y=-1\)

\(H=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)\)

Đặt \(t=x^3+3x\) ta có:

\(t\left(t+2\right)=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\). Dấu "=" xảy ra khi \(t=-1\Rightarrow x^2+3x=-1\Rightarrow x_{1,2}=\frac{-3\pm\sqrt{5}}{2}\)

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

a) xác định khi x khác +-1

b)

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

\(A=\left(\frac{\left(2x^2+3x+1\right)+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x+1}\)

\(A=\frac{x^2+5x+8}{\left(x+1\right)^2}=1+\frac{3\left(x+1\right)+4}{\left(x+1\right)^2}\)

c)

GTNN \(B=\frac{3y+4}{y^2}\ge-\frac{9}{16}\)

GTNN \(A=\frac{7}{16}\)

\(A=2.\left(x^2-2.\frac{1}{4}.x+\frac{1}{16}\right)-\frac{9}{8}=2.\left(x-\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

100 + 800 = 900

tk mình đi mình tk lại cho

tổng là:

    100 +800 = 900

            Đáp số:900