2. 

\(\left(x\pm\dfrac{4}{3}\right)^{2000}\ge0\forall x\\ =>B\ge105\)

Dấu ''='' xảy ra `<=>` \(x=\mp\dfrac{4}{3}\)

Vậy GTNN `=105<=>` \(x=\mp\dfrac{4}{3}\)

3. \(\left(x+\dfrac{3}{8}\right)^{442}\ge0\forall x\\ =>-\left(x+\dfrac{3}{8}\right)^{442}\le0\\ =>B\le-\dfrac{2}{3}\)

Dấu ''='' xảy ra `<=>x=-3/8`

Vậy GTLN `=-2/3<=>x=-3/8`

x/y=1/2 => x=1/2y

=> x+3y/x-3y = 1/2y+3y / 1/2y-3y = 3,5y/-2,5y = 3,5/-2,5 = 0,7/-0,5

`x/y =1/2 -> 2x =y`
Khi đó :
`(x+3y)/(x-3y) = (x + 3 . 2x)/(x - 3 . 2x) = (7x)/(-5x) = -7/5`

\(x-\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\right)=\dfrac{1}{7}-\dfrac{1}{3}\)

\(\Leftrightarrow x-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{1}{7}-\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{7}-\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{1}{5}\)

 x\(=\dfrac{1}{5}\)

\(x-\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\right)=\dfrac{1}{7}-\dfrac{1}{3}\)

\(\Rightarrow x-\dfrac{41}{105}=-\dfrac{4}{21}\)

\(\Rightarrow x=-\dfrac{4}{21}+\dfrac{41}{105}\)

\(\Rightarrow x=\dfrac{1}{5}\)

(0,5)²+2x=(0,7)²

=>0,25+2x=0,49

=>2x=0,49-0,25

=>2x=0,24

=>x=0,24:2

=>x=0,12

Vậy x=0,12

(0,5)² + 2 . x = (0,7)²

0,25 + 2 . x = 0,49

2 . x = 0,49 - 0,25

2 . x = 0,24

x = 0,24 : 2

x = 0,12