a) \(-\left|2x-4\right|+2016\)

Vì: \(\left|2x-4\right|\ge0\) , với mọi x

=> \(-\left|2x-4\right|\le0\)

=> \(-\left|2x-4\right|+2016\le2016\)

Vậy GTLN của bt đã cho la 2016 khi \(2x-4=0\Leftrightarrow x=2\)

b) \(1981+\left|x-4\right|\)

Vì: \(\left|x-4\right|\ge0\) , với mọi x

=> \(1981+\left|x-4\right|\ge1981\)

Vậy GTNN của bt đã cho là 1981 khi \(x-4=0\Leftrightarrow x=4\)

Ta có :

\(3A=\frac{3x^2}{x^4+x^2+1}=\frac{x^4+x^2+1-x^4+2x^2-1}{x^4+x^2+1}=\frac{\left(x^4+x^2+1\right)-\left(x^2-1\right)^2}{x^4+x^2+1}\)

\(=1-\frac{\left(x^2-1\right)^2}{x^4+x^2+1}\le1\)

\(\Leftrightarrow3A\le1\Rightarrow A\le\frac{1}{3}\)có GTLN là \(\frac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\pm1\)

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 \(x=-\dfrac{2}{3}\)

\(A=x^2-5x+1=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{21}{4}=\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\)

nên \(\left(x-\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Vậy \(Min_{x^2-5x+1}=-\frac{21}{4}\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

\(B=1-x^2+3x=-\left(x^2-3x-1\right)=-\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)Vì \(\left(x-\frac{3}{2}\right)^2\ge0\)

nên \(-\left(x-\frac{3}{2}\right)^2\le0\)

do đó \(-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\)

Vậy \(Max_{1-x^2+3x}=\frac{13}{4}\)khi \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Ta có : \(M=\dfrac{1}{a+b^2}+\dfrac{1}{b+a^2}=\dfrac{a+1}{\left(a+b^2\right)\left(a+1\right)}+\dfrac{b+1}{\left(b+1\right)\left(b+a^2\right)}\le\dfrac{a+1}{\left(a+b\right)^2}+\dfrac{b+1}{\left(a+b\right)^2}=\dfrac{1}{a+b}+\dfrac{2}{\left(a+b\right)^2}\le\dfrac{1}{2}+\dfrac{2}{4}=1\)đẳng thức xả ra khi và chỉ khi a=b=1. Do đó GTLN của M là 1.

Ta có: \(-\left|1,5-x\right|\le0\forall x\)

\(\Rightarrow-\left|1,5-x\right|-2\le-2\forall x\)

Dấu \("="\) xảy ra khi \(\left|1,5-x\right|=0\)

\(\Rightarrow1,5-x=0\Rightarrow x=1,5\)

Vậy \(Min_A=-2\) khi \(x=1,5.\)

-2

-2

\(A=\dfrac{2}{x+\sqrt{x}+1}\)

Ta có : \(x+\sqrt{x}+1=\left(x+2.\dfrac{1}{2}.\sqrt{x}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}\le\dfrac{2}{\dfrac{3}{4}}=\dfrac{2.4}{3}=\dfrac{8}{3}\)

Vậy GTLN của A là \(\dfrac{8}{3}\). Dấu "=" xảy ra khi và chỉ khi \(x=-\dfrac{1}{2}\)

Mà x > 0, nên trường hợp này ta không chấp nhận .

Ta có : Vì x > 0 , \(\Rightarrow x+\sqrt{x}+1\ge1\)

Vậy giá trị nhỏ nhất là \(1\). Dấu "=" xảy ra khi và chỉ khi \(x=1.\)