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I zì:vv
a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)
Vậy MinA=10 khi \(x=-\dfrac{1}{2}\)
b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)
\(=-\left(x+4\right)^2+21\le21\forall x\)
Vậy MaxB=21 khi x=-4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A=4x-x^2+3\)
\(A=-\left(x^2-4x-3\right)\)
\(A=-\left(x^2-4x+4\right)+7\)
\(A=-\left(x-2\right)^2+7\)
Mà: \(-\left(x-2\right)^2\le0\forall x\) nên: \(A=-\left(x-2\right)^2+7\le7\)
Dấu "=" xảy ra:
\(-\left(x-2\right)^2+7=7\)
\(\Rightarrow x=2\)
Vậy: \(A_{max}=7\) khi \(x=2\)
2) \(B=x-x^2\)
\(B=-x^2+x\)
\(B=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\) nên \(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu "=" xảy ra:
\(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy: \(B_{max}=\dfrac{1}{4}\) với \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a a= -4x^2-4x+3
để a lơn nhất thì
\(-4x^2-4x\ge0\\ \Leftrightarrow4x^2+4x\le0\\ \Leftrightarrow x^2+x\le0\\ \Leftrightarrow x\left(x+1\right)\le0\)
a lớn nhất thì x(x+1)=0 khi đó \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)thì a đạt giá trị lớn nhất là 3
B=\(\frac{1}{x^2}-6x+11\)đúng ko ?