1/4 x 1/5 x X = 1/2

                X = 1/2 : 1/5 : 1/4

                X = 10

1/4x1/5xX=1/2

            X=1/2:1/4:1/5

            X=10

    \(\frac{1}{4}\times\frac{1}{5}\times x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}\div\left(\frac{1}{4}\times\frac{1}{5}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\div\frac{1}{20}\)

\(\Leftrightarrow x=10\)

GTNN=2 <=> x=2

cần gấp

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:

\(A=\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{4}{x^2+y^2}=\frac{4}{20}=\frac{1}{5}\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}x^2+y^2=20\\x^2=y^2\end{matrix}\right.\)\(\Rightarrow x=y=\pm\sqrt{10}\)

Vậy \(Min_A=\frac{1}{5}\) khi \(x=y=\pm\sqrt{10}\)

\(A=x-2\sqrt{x}\left(\sqrt{y}+1\right)+\left(\sqrt{y}+1\right)^2+\left(3y+1-\left(\sqrt{y}+1\right)^2\right)\)

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

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

Amin= -1/2  khi  y=1/4; x=9/4

\(A=x^2-3x+1\)

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

\(=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+\frac{4}{4}\)

\(=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\) \(\forall\) \(x\) \(\Rightarrow\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\) \(\forall\) \(x\)

Vậy GTNN của A là \(-\frac{5}{4}\) tại \(x=\frac{3}{2}\)

A=[(x-1)(x+6)][(x+2)(x+3)]

=(x²+5x-6)(x²+5x+6)

=(x²+5x)²-36

Ta thấy (x²+5x)²  >=0 nên (x²+5x)²-36 >=-36

Vậy GTNN của A là -36