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\(2x^2+10x-1\)
\(=2\left(x^2+5x-\frac{1}{2}\right)\)
\(=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{27}{4}\right)\)
\(=2\left(\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right)\)
\(=\frac{-27}{2}-2\left(x+\frac{5}{2}\right)^2\le\frac{-27}{2}\)
\(MinB=\frac{-27}{2}\Leftrightarrow x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)
\(25x^2+16y^2=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)
\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)
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)
\(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)
\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)
\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)
4.
= x\(^2\)-2.\(\dfrac{5}{2}\)x+\(\dfrac{25}{4}\)-\(\dfrac{13}{4}\)
= (x-\(\dfrac{5}{2}\))\(^2\)-\(\dfrac{13}{4}\)lớn hơn hoặc bằng -\(\dfrac{13}{4}\) với mọi x
=> min= -\(\dfrac{13}{4}\) <=> x = 5/2
5.
= 2( x\(^2\)-\(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\))
=2( x\(^2\)-2.\(\dfrac{5}{4}\)+\(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))
=2( x-\(\dfrac{5}{4}\))\(^2\)-\(\dfrac{27}{2}\) lớn hơn hoặc bằng -27/2 với mọi x
vậy min = -\(\dfrac{27}{2}\) <=> x= 5/4