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\(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)
Mặt khác, ta cũng có: \(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x-4y\right)^2=10xy\)
Do đó:
\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)
Vậy, \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)
1)
\(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)
\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)
\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)
\(B=9x-3x^2=-3\times\left(x^2-2\times x\times\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right)=-3\times\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\)
\(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
\(-3\times\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\le\frac{27}{4}\)
Vậy Max B = \(\frac{27}{4}\) khi x = \(\frac{3}{2}\)
\(B=9x-3x^2\)
\(=3\left(x^2-2x\right)\)
\(=3\left(x^2-2x+1-1\right)\)
\(=-3+3\left(x-1\right)^2\ge-3\)
Max \(B=-3\Leftrightarrow x-1=0\Rightarrow x=1\)
\(A=\)\(36x^2\)\(+\)\(24x\)\(+7\)
\(\Leftrightarrow\)\(A=36x^2+24x+4+3\)
\(\Leftrightarrow\)\(A=\left(6x+2\right)^2+3\)
Vì \(\left(6x+2\right)^2\)\(\ge0\) nên \(A\ge3\)
\(\Rightarrow GTNN\)của \(A\)là \(3\) khi \(\left(6x+2\right)^2=0.\)
\(\Leftrightarrow\)\(x=-\frac{1}{3}\)
Vậy GTNN của \(A\)là \(3\)khi \(x=-\frac{1}{3}\)
