\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)

`A=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`

`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

lập bảng xét dấu đi

a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)

\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2

\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)

\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1

\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1

\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2

d)D=\(x-2\sqrt{x+2}=\left(x+2\right)-2\sqrt{x+2}+1-3\)

    \(=\left(\sqrt{x+2}-1\right)^2-3\ge-3\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4