Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

\(M=2x+\sqrt{5-x^2}\)

\(\Leftrightarrow M-2x=\sqrt{5-x^2}\)

\(\Leftrightarrow M^2-4Mx+4x^2=5-x^2\)

\(\Leftrightarrow5x^2-4Mx+M^2-5=0\)

Để PT theo nghiệm x có nghiệm thì

\(\Delta'=4M^2-5.\left(M^2-5\right)\ge0\)

\(\Leftrightarrow M^2\le25\)

\(\Leftrightarrow-5\le M\le5\)

Max đúng

Min sai rồi

DK \(x\ge-\sqrt{5}\)

=> M \(\ge-2\sqrt{5}\)

\(M=2x+\sqrt{5-x^2}\)

\(\Leftrightarrow M-2x=\sqrt{5-x^2}\)

\(\Leftrightarrow M^2-4Mx+4x^2=5-x^2\)

\(\Leftrightarrow5x^2-4Mx+M^2-5=0\)

Để PT theo nghiệm x có nghiệm thì

\(\Delta=4M^2-5.\left(M^2-5\right)\ge0\)

\(\Leftrightarrow M\le25\)

\(\Leftrightarrow-5\le M\le5\)

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

\(\Rightarrow P=\sqrt{x^2-2x+5}\ge\sqrt{4}=2\)

\(minP=2\Leftrightarrow x=1\)

Với các số thực không âm a; b ta luôn có BĐT sau:

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) (bình phương 2 vế được \(2\sqrt{ab}\ge0\) luôn đúng)

Áp dụng:

a. 

\(A\ge\sqrt{x-4+5-x}=1\)

\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(A\le\sqrt{\left(1+1\right)\left(x-4+5-x\right)}=\sqrt{2}\) (Bunhiacopxki)

\(A_{max}=\sqrt{2}\) khi \(x-4=5-x\Leftrightarrow x=\dfrac{9}{2}\)

b.

\(B\ge\sqrt{3-2x+3x+4}=\sqrt{x+7}=\sqrt{\dfrac{1}{3}\left(3x+4\right)+\dfrac{17}{3}}\ge\sqrt{\dfrac{17}{3}}=\dfrac{\sqrt{51}}{3}\)

\(B_{min}=\dfrac{\sqrt{51}}{3}\) khi \(x=-\dfrac{4}{3}\)

\(B=\sqrt{3-2x}+\sqrt{\dfrac{3}{2}}.\sqrt{2x+\dfrac{8}{3}}\le\sqrt{\left(1+\dfrac{3}{2}\right)\left(3-2x+2x+\dfrac{8}{3}\right)}=\dfrac{\sqrt{510}}{6}\)

\(B_{max}=\dfrac{\sqrt{510}}{6}\) khi \(x=\dfrac{11}{30}\)

a)Ta có:A=\(\sqrt{x-4}+\sqrt{5-x}\)

        =>A²=\(x-4+2\sqrt{\left(x-4\right)\left(5-x\right)}+5-x\)

        =>A²= 1+\(2\sqrt{\left(x-4\right)\left(5-x\right)}\ge1\)

        =>A\(\ge\)1

Dấu '=' xảy ra <=> x=4 hoặc x=5

Vậy,Min A=1 <=>x=4 hoặc x=5

Còn câu b tương tự nhé

a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)

\(=-\left(x+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTLN là 4 khi x = -1 

b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)

\(=-\left(2x-1\right)^2-2\le-2\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTLN B là -2 khi x = 1/2 

c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)

\(=-\left(x-1\right)^2-14\le-14\)

Vâỵ GTLN C là -14 khi x = 1

Bài 8 : 

b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

Dấu ''='' xảy ra khi x = 3

Vậy GTNN B là 2 khi x = 3 

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

Dấu ''='' xảy ra khi x = 1/2 

Vậy ...

c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)

Dấu ''='' xảy ra khi x = 6

Vậy ...

2:

|x+4|>=0

=>-|x+4|<=0

=>B<=11

Dấu = xảy ra khi x=-4

Đặt \(\sqrt{x^2+4}=a\ge2\)

\(\Rightarrow x^2=a^2-4\)

\(\Rightarrow A=\dfrac{2\left(a^2-4\right)+3}{a+2}=\dfrac{2a^2-5}{a+2}=2a-4+\dfrac{3}{a+2}\)

\(A=\dfrac{3\left(a+2\right)}{16}+\dfrac{3}{a+2}+\dfrac{29}{16}a-\dfrac{35}{8}\ge2\sqrt{\dfrac{9\left(a+2\right)}{16\left(a+2\right)}}+\dfrac{29}{16}.2-\dfrac{35}{8}=\dfrac{3}{4}\)

\(A_{min}=\dfrac{3}{4}\) khi \(a=2\Rightarrow x=0\)