\(B=2x^2+10x-1\)

=> \(B=2\left(x^2+5x\right)-1\)

=> \(B=2\left(x^2+2.x\frac{5}{2}+\frac{25}{4}\right)-\frac{27}{2}\)

=> \(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)

Có \(2\left(x+\frac{5}{2}\right)^2\ge0\)với mọi x

=> \(2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\)

Dấu "=" xảy ra <=> \(\left(x+\frac{5}{2}\right)^2=0\)<=> \(x+\frac{5}{2}=0\)<=> \(x=\frac{-5}{2}\)

KL: B_min = \(\frac{-27}{2}\)<=> \(x=\frac{-5}{2}\)

\(C=5x-x^2\)

=> \(C=-\left(x^2-5x\right)\)

=> \(C=-\left(x^2-2.x.\frac{5}{2}+\frac{25}{4}\right)+\frac{25}{4}\)

=> \(C=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)

Có \(\left(x-\frac{5}{2}\right)^2\ge0\)với mọi x

=> \(-\left(x-\frac{5}{2}\right)^2\le0\)

=> \(C=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra <=> \(\left(x-\frac{5}{2}\right)^2=0\)<=> \(x-\frac{5}{2}=0\)<=> \(x=\frac{5}{2}\)

KL: C_max = \(\frac{25}{4}\)<=> \(x=\frac{5}{2}\)

B=2x²+10x-1=2(x²+5x-1/2)=2(x²+2*5/2*x+25/4-27/4)=2[x²+2*5/2*x+(5/2)²]-27/2=2(x+5/2)²-27/2

Ta có: (x+5/2)^2>=0(với mọi x)

=> 2(x+5/2)^2>=0(với mọi x)

=> 2(x+5/2)^2-27/2>=-27/2(với mọi x)

hay B>=-27/2( với mọi x)

Do đó, GTNN của B là -27/2 khi:

x+5/2=0

x=-5/2

Vậy GTNN của B là -27/2 khi x=-5/2

C=5x-x^2=-x^2+5x=-x^2+2*5/2*x-25/4+25/4=-[x^2-2*5/2*x+(5/2)^2]+25/4=-(x-5/2)^2+25/4

Ta có: (x-5/2)^2>=0(với mọi x)

=>-(x-5/2)^2<=0(với mọi x)

=> -(x-5/2)^2+25/4<=25/4(với mọi x) hay C<=25/4(với mọi x)

Do đó, GTLN của C là 25/4 khi: x-5/2=0

                                              x=5/2

Vậy GTLN của C là 25/4 tại x=5/2

GTNN A= 0

GTNN B= -1

GTLN C = 0,5

GTLN D = 3

Để : \(A=3,7\left|4,3-x\right|min\)

Thì :\(\left|4,3-x\right|\)Phải min

Ta có :\(\left|4,3-x\right|\ge0\)

\(\Rightarrow\left|4,3-x\right|min=0\)

\(\Rightarrow4,3-x=0\Rightarrow x=4,3\)

\(\Rightarrow Amin=3,7X4.3=15.91\)

Theo đề bài ta có: \(A=\left(2x+\frac{1}{3}\right)^4-1\)

Nhận xét: \(\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x+\frac{1}{3}\right)^4=0\Rightarrow x=\frac{-1}{6}\)

Vậy \(minA=-1\Leftrightarrow x=\frac{-1}{6}\)

Theo đề bài, ta có:

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

Nhận xét: 

\(\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\forall x\)

Dấu\("="\)xảy ra khi \(\Leftrightarrow\left(2x+\frac{1}{3}\right)^4=0\Rightarrow x=\frac{-1}{6}\)

Vậy \(A=-1\Leftrightarrow x=\frac{-1}{6}\)

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

Mà \(\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2x+\frac{1}{3}\right)^4=0\)

\(\Leftrightarrow x=\frac{-1}{6}\)

Vậy \(Min_A=-1\Leftrightarrow x=\frac{-1}{6}\)

a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)