Mình tách thành hai phần nhìn cho dễ hiểu nhé !

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)

+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x+x-9-x+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)

\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)

a) \(-ĐKXĐ:x\ne\pm2;1\)

Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)

\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)

b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)

\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)

Vậy với mọi x thỏa mãn x>1 thì A > 0

c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)

\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy x = -1;-2

Xét mẫu (x-2)²+(x-y)⁴+3

R đạt GTLN khi (x-2)²+(x-y)⁴+3 nhỏ nhất

Ta có \(\left(x-2\right)^2\ge0\)

\(\left(x-y\right)^4\ge0\)

=>(x-2)²+(x-y)⁴+3\(\ge3\)

Vậy mẫu số đạt GTNN là 3 khi x=y=2

Khi đó GTLN của R là 2013/3

Vì \(\left(x-2\right)^2\ge0\forall x\in R\)

     \(\left(x-y\right)^4\ge0\forall x;y\in R\)

\(\Rightarrow\left(x-2\right)^2+\left(x-y\right)^2+3\ge3\forall x;y\in R\)

 Để biểu thức\(R_{max}\Leftrightarrow\)\(\left(x-2\right)^2+\left(x-y\right)^4+3=3\Rightarrow\left(x-2\right)^2=\left(x-y\right)^4=0\)

Ta có \(:\)\(\left(x-2\right)^2=0\Rightarrow x=0+2=2\)

Thay \(x=2\)vào \(\left(x-y\right)^4=0\)ta có \(:\)

\(\left(x-y\right)^4=\left(2-y\right)^4=0\Rightarrow y=2-0=2\)

\(\Rightarrow R_{max}=\frac{2013}{\left(2-2\right)^2+\left(2-2\right)^2+3}=\frac{2013}{3}\)

           Vậy GTLN của \(R=\frac{2013}{3}\)tại \(x=2;y=2\)

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

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