\(M=x^2-8x+5\)
\(\Leftrightarrow M=x^2-8x+16-11\)
\(\Leftrightarrow M=\left(x-4\right)^2-11\ge-11\)
Min M = -11
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
\(N=-3x-6x-9\)
\(\Leftrightarrow N=-9x-9\le-9\)
Max N = -9
\(\Leftrightarrow x=0\)
a) Ta có : M = x2 - 8x + 5 = x2 - 8x + 16 - 17 = (x - 4)2 - 17 \(\ge\)-17
Dấu "=" xảy ra <=> x - 4 = 0 => x = 4
Sorry
a) M = x2 - 8x + 5
= x2 - 8x + 16 - 11
=> (x - 4)2 - 11 \(\ge\)-11
Dấu "=" xảy ra <=> x - 4 = 0 => x = 4
Vậy Min M = -11 <=> x = 4
a,\(M=x^2-8x+16-11=\left(x-4\right)^2-11\)
Ta có : \(\left(x-4\right)^2\ge0< = >\left(x-4\right)^2-11\ge-11\)
Dấu = xảy ra \(< =>x-4=0< =>x=4\)
Vậy\(Min_M=-11\)khi \(x=4\)
b, \(N=-3x^2-6x-9=\left(-x^2-2x-1\right)+\left(-x^2-2x-1\right)+\left(-x^2-2x-1\right)-6\)
\(=-\left(x+1\right)^2-\left(x+1\right)^2-\left(x+1\right)^2-6\)
\(=-\left[3\left(x+1\right)^2+6\right]\)
Ta có : \(3\left(x+1\right)^2\ge0< =>3\left(x+1\right)^2+6\ge6\)
\(< =>-\left[3\left(x+1\right)^2+6\right]\le-6\)
Dấu = xảy ra \(< =>3\left(x+1\right)^2=0< =>x=-1\)
Vậy \(Max_N=-6\)khi \(x=-1\)
\(Q=x^2+5y^2+2xy-2y+2020\)
\(\Leftrightarrow Q=x^2+2xy+y^2+4y^2-2.2y.\frac{1}{2}+\frac{1}{4}+\frac{8079}{4}\)
\(\Leftrightarrow Q=\left(x+y\right)^2+\left(2y+\frac{1}{2}\right)^2+\frac{8079}{4}\ge\frac{8079}{4}\)
Min Q= \(\frac{8079}{4}\)
\(\Leftrightarrow\hept{\begin{cases}x+y=0\\2y+\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\y=\frac{-1}{4}\end{cases}\Leftrightarrow\hept{\begin{cases}x-\frac{1}{4}=0\\y=\frac{-1}{4}\end{cases}\Leftrightarrow}}\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{-1}{4}\end{cases}}\)
dcv_new Phần b bạn chép lộn đề rùi kìa
đề như vậy sẽ không có giá trị lớn nhất hay nhỏ nhất
x có thể âm hoặc dương tùy ý
Bài giải
\(M=x^2-8x+5=x^2-2\cdot4x+16-11=\left(x-4\right)^2-11\ge-11\)
Dấu " = " xảy ra khi \(\left(x-4\right)^2=0\text{ }\Rightarrow\text{ }x-4=0\text{ }\Rightarrow\text{ }x=4\)
Vậy \(Min_M=-11\text{ khi }x=4\)
\(N=-3x-6x-9=-9x-9=-9\left(x+1\right)\le0\)
Dấu " = " xảy ra khi \(-9\left(x+1\right)=0\text{ }\Rightarrow\text{ }x+1=0\text{ }\Rightarrow\text{ }x=-1\)
\(\Rightarrow\text{ }Max_N=0\text{ khi }x=-1\)
M = x2 - 8x + 5
= x2 - 8x + 16 - 11
= ( x - 4 )2 - 11
\(\left(x-4\right)^2\ge0\forall x\Rightarrow\left(x-4\right)^2-11\ge-11\)
Dấu " = " xảy ra <=> x - 4 = 0 => x = 4
Vậy MinM = -11, đạt được khi x = 4
N = -3x2 - 6x - 9
= -3( x2 + 2x + 1 ) - 6
= -3( x + 1 )2 - 6
\(-3\left(x+1\right)^2\le0\forall x\Rightarrow-3\left(x+1\right)^2-6\le-6\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy MaxN = -6, đạt được khi x = -1
Q = x2 + 5y2 + 2xy - 2y + 2020
= ( x2 + 2xy + y2 ) + ( 4y2 - 2y + 1/4 ) + 8079/4
= ( x + y )2 + ( 2y - 1/2 )2 + 8079/4
\(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x,y\\\left(2y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+y\right)^2+\left(2y-\frac{1}{2}\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x+y\right)^2+\left(2y-\frac{1}{2}\right)^2+\frac{8079}{4}\ge\frac{8079}{4}\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\2y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{4}\\y=\frac{1}{4}\end{cases}}\)
Vậy MinQ = 8079/4, đạt được khi x = -1/4, y = 1/4
