thì A=\(\left|3-x\right|+\left|x-2\right|\ge\left|3-x+x-2\right|=1\) (bất đẳng thức về dâu giá trị tuyệt đối)

dấu = xảy ra <=> tích của chúng = nhau

Tổng các hệ số của đa thức \(A\left(x\right)\) bất kì bằng giá trị của đa thức đó tại \(x=1\).

Thay \(x=1\) vào đa thức \(A\left(x\right)\) ta có:

\(A\left(1\right)=\left(3-4+1\right)^{2004}.\left(3+4+1\right)^{2005}=0\)

1) \(B=-7x^2+9\)

Do \(x^2\ge0\forall x\Rightarrow-7x^2\le0\forall x\)

\(\Rightarrow B=-7x^2+9\le9\)

\(maxB=9\Leftrightarrow x=0\)

2) \(C=2-\left(3x-4\right)^4\)

Do \(\left(3x-4\right)^4\ge0\forall x\Rightarrow-\left(3x-4\right)^4\le0\forall x\)

\(\Rightarrow C=2-\left(3x-4\right)^4\le2\)

\(maxC=2\Leftrightarrow x=\dfrac{4}{3}\)

3) \(D=\dfrac{1}{2}x^2+3\)

Do \(\dfrac{1}{2}x^2\ge0\forall x\Rightarrow D=\dfrac{1}{2}x^2+3\ge3\)

\(minD=3\Leftrightarrow x=0\)

4) \(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{-x^2+5}\)

Do \(x^2\ge0\forall x\Rightarrow-x^2+5\le5\forall x\)

\(\Rightarrow E=\dfrac{2016}{-x^2+5}\ge\dfrac{2016}{5}\)

\(minE=\dfrac{2016}{5}\Leftrightarrow x=0\)

\(B=-7x^2+9\)

Vì \(-7x^2\le0\forall x\)

\(\Rightarrow-7x^2+9\le9\forall x\)

\(\Rightarrow B_{max}=9\Leftrightarrow-7x^2=0\Leftrightarrow x=0\)

\(C=2-\left(3x-4\right)^4\)

Vì \(-\left(3x-4\right)^4\le0\forall x\)

\(\Rightarrow-\left(3x-4\right)^4+2\le2\forall x\)

\(\Rightarrow C_{max}=2\Leftrightarrow-\left(3x-4\right)^4=0\Leftrightarrow x=\dfrac{4}{3}\)

Nếu tìm GTLN thì câu \(d\) là \(D=-\dfrac{1}{2}x^2+3\)

Vì \(-\dfrac{1}{2}x^2\le0\forall x\)

\(\Rightarrow-\dfrac{1}{2}x^2+3\le3\forall x\)

\(\Rightarrow D_{max}=3\Leftrightarrow-\dfrac{1}{2}x^2=0\Leftrightarrow x=0\)

\(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{5-x^2}\)

Vì \(x^2\ge0\forall x\)

\(\Rightarrow5-x^2\le5\forall x\)

\(\Rightarrow E_{min}=5\Leftrightarrow x=\dfrac{2016}{5}\)

a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)

\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)

Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)

Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).

\(---\)

b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)

\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)

Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).

\(---\)

c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)

\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)

\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)

Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)

Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).

a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)

\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)

Dấu "=" xảy ra khi:

\(\dfrac{2}{5}-x=0\)

\(\Rightarrow x=\dfrac{2}{5}\)

Vậy: ... 

b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)

\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)

Dấu "=" xảy ra:

\(x+\dfrac{2}{3}=0\)

\(\Rightarrow x=-\dfrac{2}{3}\)

Vậy: ...

c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)

\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)

Dấu "=" xảy ra:

\(\dfrac{7}{4}-x=0\)

\(\Rightarrow x=\dfrac{7}{4}\)

Vậy: ...

\(A=0,5-\left|x-3,5\right|\le0,5\\ A_{max}=0,5\Leftrightarrow x-3,5=0\Leftrightarrow x=3,5\\ B=-\left|1,4-x\right|2=-2\left|1,4-x\right|\le0\\ B_{min}=0\Leftrightarrow1,4-x=0\Leftrightarrow x=1,4\)