\(\text{Đặt: }x^3=a\Rightarrow B=16a-a^2=-64+16a-a^2+64=-\left(a-8\right)^2+64\le64.\text{ Do đó: }B_{max}=64.\text{Dấu }"="\text{ xảy ra khi: }a=8\text{ hay: }x=2\)

\(C=-x^2-y^2+xy+2x+2y\Leftrightarrow2C=-2x^2-2y^2+2xy+4x+4y=-\left(x^2-2xy+y^2\right)-\left(x^2-4x+4\right)-\left(y^2-4y+4\right)+8=-\left[\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\right]+8\le8\)\(\Rightarrow C\le4\)

Dấu đẳng thức xảy ra <=> x = y = 2

Vậy \(MaxC=4\Leftrightarrow x=y=2\)

`[2x]/[x+3]-[x-1]/[3-x]=[3x^2+1]/[x^2-9]`       `ĐK: x \ne +-3`

`<=>[2x(x-3)+(x-1)(x+3)]/[(x-3)(x+3)]=[3x^2+1]/[(x-3)(x+3)]`

   `=>2x^2-6x+x^2+3x-x-3=3x^2+1`

`<=>-4x=4`

`<=>x=-1` (t/m)

Vậy `S={-1}`

b, -(2x-1)²+10I2x-1I+2018

Vì :

(2x-1)² >= 0  với mọi x

=> -(2x-1)² =< -0 với mọi x    ₁

I2x-1I >= 0 với mọi x

=> 10I2x-1I >= 0 với mọi x    ₂

Từ (1) và (2) : 

=> -(2x-1)²+10I2x-1I =< -0 với mọi x

=> -(2x-1)²+10I2x-1I +2018 =< -0+2018  với mọi x

=> -(2x-1)²+10I2x-1I +2018 =< - 2018       với mọi x   

=> GTLN là -2018

Vậy GTLN là -2018 .

A=3(x^2+2/3x-1)

=3(x^2+2*x*1/3+1/9-10/9)

=3(x+1/3)^2-10/3>=-10/3

Dấu = xảy ra khi x=-1/3

\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)

Dấu = xảy ra khi x=-1/2

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)