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 \(x=-\dfrac{2}{3}\)

Ta có:\(3x^2+6x+9=3\left(x^2+2x+3\right)=3\left[\left(x^2+2x+1\right)+2\right]=3\left[\left(x+1\right)^2+2\right]\)

\(=3\left(x+1\right)^2+6\)

Vì \(3\left(x+1\right)^2\ge0\forall x\Rightarrow3\left(x+1\right)^2+6\ge6\forall x\)

Dấu "=" xảy ra khi: \(3\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy GTNN của \(3x^2+6x+9\) là 6 khi x = -1.

\(3x^{2}-6x+9\)

\(\Leftrightarrow\)\(3(x^{2}-2x+3)\)

\(\Leftrightarrow\)\(3(x^{2}-2x+1)+2\)

\(\Leftrightarrow\)\(3(x-1)^{2}+2\)

GTNN = 2. Dấu "=" xảy ra khi \(x=1\)

\(A=\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)+\left(x^2+2x+1\right)+1\\ A=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2+\left(x+1\right)^2+1\ge1\\ A_{min}=1\Leftrightarrow x=y=z=-1\)

A=3(x^2+2/3x-1)

=3(x^2+2*x*1/3+1/9-10/9)

=3(x+1/3)^2-10/3>=-10/3

Dấu = xảy ra khi x=-1/3

\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)

Dấu = xảy ra khi x=-1/2

Chọn A

Xét hàm số y = x 3 + 3 x 2 - 9 x + 1  trên đoạn [-4;4].

Ta có:

y(1) = -4, y(-3) = 28; y(4) = 77; y(-4) = 21

GTNN của hàm số y = x 3 - 9 x + 1  trên đoạn [-4;4] là -4 khi x= 1

\(y=x^3-3x^2-9x+35\)

\(y'=3x^2-6x-9\)

\(y'=0\Leftrightarrow3x^2-6x-9=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

\(y\left(-4\right)=-41;y\left(-1\right)=40;y\left(3\right)=8;y\left(4\right)=52\)

\(\Rightarrow y_{max}=y\left(4\right)=52;y_{min}=y\left(-4\right)=-41\) trên đoạn \(\left[-4;4\right]\)

\(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4\)

Kl: Min_A = 4

\(B=x^2-x+1=\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

kl:.......

\(C=5x^2+5x+1=5\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+1-\dfrac{5}{4}=5\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

kl:.......

\(D=3x^2+4x+2=3\left(x^2+2\cdot\dfrac{2}{3}x+\dfrac{4}{9}\right)+2-\dfrac{4}{3}=3\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)

kl:......

\(E=\dfrac{1}{2}\cdot x^2+x-1=\dfrac{1}{2}\left(x^2+2x+1\right)-1-\dfrac{1}{2}=\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)

kl:............

\(F=\dfrac{1}{9}x^2+3x+2=\dfrac{1}{3}\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+2-\dfrac{1}{12}=\dfrac{1}{3}\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\)

kl:..........