\(A=2\left|x-5\right|-2015\ge-2015\)

\(Min_A=-2015\Leftrightarrow x=5\)

\(B=205-\left|3x-5\right|\le205\)

\(Max_B=205\Leftrightarrow x=\frac{5}{3}\)

có cách làm củ thể hơn k bạn

2/

a, \(A=2x^2+6x-5=2\left(x^2+3x-\frac{5}{2}\right)=2\left(x^2+2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{19}{4}\right)=2\left[\left(x+\frac{3}{2}\right)^2-\frac{19}{4}\right]=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

Dấu "=" xảy ra khi x=-3/2

Vậy Amin=-19/2 khi x=-3/2

b,bài này phải tìm min 

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

Vì \(\left(x-2\right)^2\ge0\Rightarrow B=\left(x-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi x = 2

Vậy Bmin=4 khi x=2

Bài 2)Ta có:

\(2x^2+6x-5\)

\(=2x^2+6x+\frac{9}{2}-\frac{19}{2}\)

\(=2\left(x^2+3x+\frac{9}{4}\right)-\frac{19}{2}\)

\(=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(P\le\sqrt{2\left(3x-5+7-3x\right)}=2\)

\(P_{max}=2\) khi \(3x-5=7-3x\Rightarrow x=2\)

\(A=2\left(x-1\right)+\dfrac{9}{x-1}+2\ge2\sqrt{\dfrac{18\left(x-1\right)}{x-1}}+2=6\sqrt{2}+2\)

\(A_{min}=6\sqrt{2}+2\) khi \(x=\dfrac{2+3\sqrt{2}}{2}\)

\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)

Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)

Dấu \("="\Leftrightarrow x=-5\)

2) ĐKXĐ:  \(1\le x\le5\)

\(B^2=\left(\sqrt{x-1}+\sqrt{5-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+5-x\right)=8\Rightarrow B\le2\sqrt{2}\)

Xảy ra đẳng thức khi và chỉ khi x = 3

b)\(\sqrt{2^3+1}\) theo mình phần b như vậy ko bít đúng ko

a)=**** 100%

b)\(\sqrt{2^3+1}\) phần b ko bít đúng ko nhưng phần a đúng ko 100%

a)=1

b)\(\sqrt{2^3+1}\) phần b ko bít đúng ko nhưng phần a đúng ko 100%

1/

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

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4