a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)

\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)

b) \(18A=1\)

<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))

<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)

<=> 18( x² - 4x + 5 ) = x³ + 6x² - 32

<=> 18x² - 72x + 90 = x³ + 6x² - 32

<=> x³ + 6x² - 32 - 18x² + 72x - 90 = 0

<=> x³ - 12x² + 72x - 122 = 0

Rồi đến đây chịu á :) 

Ý lộn == là \(\frac{x^2-2x}{x+4}\)ạ ==

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

vì \(|3x+4|\ge0\forall x\in Q\)

\(\Rightarrow1+\)\(|3x+4|\ge1\)

dấu = xảy ra <=>

3x+4=0

3x=-4

x=\(\frac{-3}{4}\)

vậy GTLN của A  lớn nhất tại x=-3/4

Vì 3>0

=>Để A đạt gtln

=>1+|3x+4| nhỏ nhất

Vì |3x+4|≥0

=>1+|3x+4|≥1

Dấu "=" xảy ra <=>3x+4=0

                         <=>3x=-4

                         <=>x=-4/3

=>Max A=3<=>x=-4/3

Để A = 3 / 1 + | 3x + 4 | đạt giá trị lớn nhất

\(\Leftrightarrow\)1 + | 3x + 4 | đạt giá trị nhỏ nhất

Ta có :

B = 1 + | 3x + 4 |

B = | 3x + 4 | + 1\(\ge\)1

Dấu " = " xảy ra \(\Leftrightarrow\)3x + 4 = 0

                            \(\Rightarrow\)x          = - 4 / 3

Min B = 1 \(\Leftrightarrow\)x = - 4 / 3

Vậy : Max A = 3 / 1 = 3 \(\Leftrightarrow\)x = - 4 / 3

Ta có : Để M=\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\left(\frac{x^2+8x+16}{32}\right)=0\)

<=> M=\(\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)=0\)

<=>M=\(\left(\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)

<=>M=\(\left(\frac{32}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)

<=>M=\(\frac{x+4}{x-4}\)

b) Thay x=\(\frac{-3}{8}\) vào M:

M=\(\frac{x+4}{x-4}=\frac{\frac{-3}{8}+4}{\frac{-3}{8}-4}=\frac{-29}{35}\)

c)Hình như sai!

d)

a)\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\)

           Vì \(-\left|x+\frac{3}{2}\right|\)\(\le\)0

        Suy ra:\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)

      Dấu = xảy ra khi \(x+\frac{3}{2}=0\)

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

Vậy Max A=\(\frac{1}{4}\) khi \(x=-\frac{3}{2}\)

b)\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)

        Vì \(-\left|x-\frac{4}{3}\right|\le0;-\left|y+\frac{1}{2}\right|\le0\)

               Suy ra:\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)

     Dấu = xảy ra khi \(x-\frac{4}{3}=0;x=\frac{4}{3}\)

                                 \(y+\frac{1}{2}=0;y=-\frac{1}{2}\)

Vậy Max B=\(\frac{5}{3}\) khi \(x=\frac{4}{3};y=-\frac{1}{2}\)

a/ Ta có ; \(\left|x+\frac{3}{2}\right|\ge0\Rightarrow-\left|x+\frac{3}{2}\right|\le0\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)

Vậy BT đạt giá trị lớn nhất bằng 1/4 khi x = -3/2

b/ \(\begin{cases}\left|x-\frac{4}{3}\right|\ge0\\\left|y+\frac{1}{2}\right|\ge0\end{cases}\) \(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\le0\\-\left|y+\frac{1}{2}\right|\le0\end{cases}\) 

\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)

\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)

Vậy BT đạt giá trị lớn nhất bằng 5/3 khi x = 4/3 , y = -1/2

a)Đặt \(A=\frac{1}{4}-\left|x+\frac{3}{2}\right|\)

Ta thấy: \(\left|x+\frac{3}{2}\right|\ge0\)

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

\(\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}-0=\frac{1}{4}\)

\(\Rightarrow A\le\frac{1}{4}\)

Dấu = khi \(x=-\frac{3}{2}\)

Vậy MaxA=\(\frac{1}{4}\Leftrightarrow x=-\frac{3}{2}\)

b)Đặt \(B=\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)

Ta thấy: \(\begin{cases}\left|x-\frac{4}{3}\right|\\\left|y+\frac{1}{2}\right|\end{cases}\ge0\)

\(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\\-\left|y+\frac{1}{2}\right|\end{cases}\)\(\le0\)

\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)

\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}-0=\frac{5}{3}\)

\(\Rightarrow B\le\frac{5}{3}\)

Dấu = khi \(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)

Vậy MaxB=\(\frac{5}{3}\Leftrightarrow\)\(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)