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a) \(N=-1-x-x^2=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\)
\(maxN=-\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)
\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)
a: Ta có: \(N=-x^2-x-1\)
\(=-\left(x^2+x+1\right)\)
\(=-\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
b: ta có: \(B=3x^2+4x-13\)
\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)
\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2x\left(x-4\right)-10=2x^2-8x-10\)
\(=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)
\(minB=-18\Leftrightarrow x=2\)
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\(A=xy+xz+2yz+2xz=x\left(y+z\right)+2z\left(x+y\right)\)
\(=x\left(6-x\right)+2z\left(6-z\right)=-x^2+6x+2\left(-z^2+6z\right)\)
\(=-\left(x-3\right)^2-2\left(z-3\right)^2+27\le27\)
\(A_{max}=27\) khi \(\left(x;y;z\right)=\left(3;0;3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=-2x^2+6x-12\)
\(=-2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{15}{2}\)
\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{15}{2}\le-\dfrac{15}{2}\)
\(maxA=-\dfrac{15}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Ta có: \(A=-2x^2+6x-12\)
\(=-2\left(x^2-3x+6\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{15}{4}\right)\)
\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{15}{2}\le-\dfrac{15}{2}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{x^2-4x+4+5}=\dfrac{1}{\left(x-2\right)^2+5}\)
Do \(\left(x-2\right)^2\ge0\) ; \(\forall x\Rightarrow\left(x-2\right)^2+5\ge5\) ; \(\forall x\)
\(\Rightarrow A\le\dfrac{1}{5}\)
\(A_{max}=\dfrac{1}{5}\) khi \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Ta có : -x2-4x+9
=-x2-4x-4+13
=-(x2+4x+4)+13
=-(x+2)2+13
=13-(x+2)2
\(\Rightarrow\)(x+2)2\(\ge\)0
Ma: 13>0 \(\Leftrightarrow\)(x+2)2\(\le\)13
Vay GTLN la 13
Dau "=" xay ra khi : x+2=0
x=-2
-x^2-4x+9=-(x^2+4x+4-13)=-(x+2)^2+13
ta co -(x+2)^2 nho hon hoac bang 0
13 lon hon 0
nen bt tren se nho hon hoac bang 13
dau = xay ra <=> x+2=0=>x=-2
vay min bt =13 tai x=-2
\(A=-x^2+4x-4-5=-\left(x-2\right)^2-5\)
Vì \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-5\le-5\)
Vậy GTLN của A là \(-5\)