Chọn A

a: ĐKXĐ: x<>-2/3

b: F=0

=>8-2x=0

=>x=4

d: F<0

=>(2x-8)/(3x+2)>0

=>x>4 hoặc x<-2/3

\(A=\frac{3-4x}{x^2+1}\Rightarrow Ax^2+4x+\left(A-3\right)=0\left(1\right)\)

Để PT (1) có nghiệm thì \(\Delta^'\ge0\)

Mà \(\Delta^'\)=2²-A(A-3)=-A²+3A+4=-(A+1)(A-4)

\(\Rightarrow-\left(A+1\right)\left(A-4\right)\ge0\Rightarrow\left(A+1\right)\left(A-4\right)\le0\Leftrightarrow-1\le A\le4\)

=>Max A=4<=>x=\(-\frac{1}{2}\)

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Lời giải:
\(x^2+3y^2+10x-14y-2xy=11\)

$\Leftrightarrow (x^2-2xy+y^2)+2y^2+10x-14y=11$

$\Leftrightarrow (x-y)^2+10(x-y)+25+(2y^2-4y+2)=38$

$\Leftrightarrow (x-y+5)^2+2(y-1)^2=38$

$\Rightarrow (x-y+5)^2=38-2(y-1)^2\leq 38$

$\Rightarrow -\sqrt{38}\leq x-y+5\leq \sqrt{38}$

$\Leftrightarrow -\sqrt{38}-5\leq x-y\leq \sqrt{38}-5$
Vậy $A_{\min}=-\sqrt{38}-5$ và $A_{\max}=\sqrt{38}-5$

\(a)\) Ta có : 

\(A=\left|x-1,35\right|\ge0\)

Dấu "=" xảy ra khi \(\left|x-1,35\right|=0\)

\(\Rightarrow\)\(x-1,35=0\)

\(\Rightarrow\)\(x=1,35\)

Vậy \(A_{min}=0\) khi \(x=1,35\)

\(b)\) Ta có : 

\(\left|2x-8\right|\ge0\)

\(\Rightarrow\)\(\frac{3}{14}-\left|2x-8\right|\le\frac{3}{14}\)

Dấu "=" xảy ra khi \(\left|2x-8\right|=0\)

\(\Rightarrow\)\(2x-8=0\)

\(\Rightarrow\)\(2x=8\)

\(\Rightarrow\)\(x=4\)

Vậy \(B_{max}=\frac{3}{14}\) khi \(x=4\)

Chúc bạn học tốt ~ 

Nhầm \(C=5-\left(x-1\right)^2\)suy ra Max C=5.

Ta có : 

\(C=4-x^2+2x\)

\(-C=x^2-2x-4\)

\(-C=\left(x^2-2x+1\right)-5\)

\(-C=\left(x-1\right)^2-5\ge5\)

\(\Rightarrow\)\(C=-\left(x-1\right)^2+5\le5\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x-1\right)^2=0\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=0\)

\(\Leftrightarrow\)\(x-1=0\)

\(\Leftrightarrow\)\(x=1\)

Vậy GTLN của \(C\) là \(5\) khi \(x=1\)

Chúc bạn học tốt ~ 

C=4-x^2+2x

   = 5-(x^2 -2x+1)

   =5-(x-1)^2

  Ta thấy (x-1)^2 lớn hơn hoặc bằng 0 với mọi x

   => C max là 5 <=> x=1

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

Vậy Min = \(\frac{1}{3}\Leftrightarrow x=-1\)

Ta có : \(\frac{x^2+x+1}{x^2-x+1}=\frac{-2\left(x^2-2x+1\right)+3\left(x^2-x+1\right)}{x^2-x+1}=\frac{-2\left(x-1\right)^2}{x^2-x+1}+3\le3\)

Vậy Max = 3  <=> x = 1

\(A=\left(1-x\right)\left(2+x\right)\le\dfrac{1}{4}\left(1-x+2+x\right)^2=\dfrac{9}{4}\)

\(A_{max}=\dfrac{9}{4}\) khi \(1-x=2+x\Leftrightarrow x=-\dfrac{1}{2}\)

\(B=\left(3-x\right)\left(2+x\right)^2=\dfrac{1}{2}\left(6-2x\right)\left(2+x\right)\left(2+x\right)\le\dfrac{1}{2}\left(\dfrac{6-2x+2+x+2+x}{3}\right)^3=\dfrac{500}{27}\)

\(B_{max}=\dfrac{500}{27}\) khi \(6-2x=2+x\Rightarrow x=\dfrac{4}{3}\)