ta có: (y^2 -25) ^4 >= 0
suy ra -2*(y^2 -25) ^4 <=0
suy ra -2*(y^2 -25) ^4+ 10 <=10
vậy GTLN là 10 khi y^2 =25 <=> y=+-5
\(A=10-2\left(y^2-25\right)^4\)
\(=10-2\left[\left(y^2-25\right)^2\right]^2\)
Ta có : \(\left(y^2-25\right)^2\ge0\forall y\)
=> \(\left[\left(y^2-25\right)^2\right]^2\ge0\forall y\)
=> \(-2\left[\left(y^2-25\right)^2\right]^2\le0\forall y\)
=> \(10-2\left[\left(y^2-25\right)^2\right]^2\le10\)
Dấu = xảy ra <=> \(10-2\left[\left(y^2-25\right)^2\right]^2=10\)
<=> \(y^2-25=0\)
<=> \(y^2=25\)
<=> \(\orbr{\begin{cases}y=5\\y=-5\end{cases}}\)
Vậy MaxA = 10 với y = \(\pm\)5
\(A=10-2\left(y^2-25\right)^4\)
Ta có \(\left(y^2-25\right)^4\ge0\forall x\Rightarrow2\left(y^2-25\right)^4\ge10\)
\(\Rightarrow10-2\left(y^2-25\right)^4\le10\)
\(\Rightarrow B\le10\)
Dấu "=" xảy ra \(\Leftrightarrow2\left(y^2-25\right)^4=0\)
\(\Leftrightarrow y^2-25=0\)
\(\Leftrightarrow\left(y-5\right)\left(y+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-5\end{cases}}\)
Vậy GTLN của B=10 đạt được khi x=5;x=-5
\(A=10-2\left(y^2-25\right)^4\)
\(=10-2\left[\left(y^2-25\right)^2\right]^2\)
Ta có:\(\left(y^2-25\right)^2\ge0\forall y\)
\(\Rightarrow\left[\left(y^2-25\right)^2\right]^2\ge\forall y\)
\(\Rightarrow-2\left[\left(y^2-25\right)^2\right]^2\le\forall y\)
\(\Rightarrow10-2\left[\left(y^2-25\right)^2\right]^2\le10\)
Dấu"=" xảy ra\(\Leftrightarrow2\left(y^2-25\right)^4=0\)
\(\Leftrightarrow y^2-25=0\)
\(\Leftrightarrow\left(y-25\right)\left(y+25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-5\end{cases}}\)
Vậy GTLN của B=10 và y=\(\pm5\)