a) Đặt \(A=x^2-2x+5\)
\(=\left(x-1\right)^2+4\)
Ta thấy \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+4\ge0+4\forall x\)
hay \(A\ge4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy Min A=4 \(\Leftrightarrow x=1\)
a , \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
Dấu " = " xảy ra khi x - 1 = 0 hay x = 1
Vậy GTNN là 4 khi x = 1 .
b , \(9-4x-x^2=-\left(x^2+4x-9\right)=-\left(x^2+4x+4-13\right)=-\left(x+2\right)^2+13=13-\left(x+2\right)^2\le13\)
Dấu " = " xảy ra khi x + 2 = 0 hay x = -2 .
Vậy GTLN là 13 khi x = -2 .
c , mik ko bt làm
b) Đặt \(B=9-4x-x^2\)
\(=-x^2-4x+9\)
\(=-x^2-4x-4+13\)
\(=-\left(x+2\right)^2+13\)
Ta thấy \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x+2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+2\right)^2+13\le0+13\forall x\)
Hay \(B\le13\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy Max B=13 \(\Leftrightarrow x=-2\)
c) Đặt C=\(-x^2-4y^2-z^2+2x-6y+10z+1975\)
\(=\left(-x^2+2x-1\right)+\left[-\left(2y\right)^2-2.2y.\frac{3}{2}-\left(\frac{3}{2}\right)^2\right]+\left(-z^2+2.z.5-5^2\right)+1975+1+5^2+\left(\frac{3}{2}\right)^2\)
\(=-\left(x-1\right)^2-\left(2y+\frac{3}{2}\right)^2-\left(z-5\right)^2+\frac{8013}{4}\)
Ta thấy \(-\left(x-1\right)^2-\left(2y+\frac{3}{2}\right)^2-\left(z-5\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-1\right)^2-\left(2y+\frac{3}{2}\right)^2-\left(z-5\right)^2+\frac{8013}{4}\le0+\frac{8013}{4}\forall x\)
Hay \(C\le\frac{8013}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2y+\frac{3}{2}=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{-3}{4}\\z=5\end{cases}}}\)
Vậy Min \(C=\frac{8013}{4}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{-3}{4}\\z=5\end{cases}}\)
