ĐKXĐ : \(x\ne\pm3\)

a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)

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

\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)

\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)

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

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

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

b) \(\left|x-5\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => ta xét x = 7

\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)

c) Để A nguyên thì 4x + 1 ⋮ 2x - 3

<=> 4x - 6 + 7 ⋮ 2x - 3

<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3

Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3

=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }

=> x thuộc { 2; 1; 5; -2 }

Vậy .....

a)   ĐKXĐ: \(x\ne\pm3\)

   \(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)

 \(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)

\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)

\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)

b)

  Có 2 trường hợp:

T.Hợp 1:

               \(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)

thay vào A ta được: A=\(-\frac{13}{8}\)

T.Hợp 2:

          \(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)

Vậy không tồn tại giá trị của A tại x=3

Vậy với x=7 thì A=-13/8

c)

      \(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)

Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)

Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .

Vậy không có giá trị nguyên nào của x để A nguyên

Câu 1:

\(P=\sqrt{a\left(a+b+c\right)+bc}+\sqrt{b\left(a+b+c\right)+ac}+\sqrt{c\left(a+b+c\right)+ab}\)

\(P=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\)

Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)

\(P\le\frac{a+b+a+c}{2}+\frac{b+a+b+c}{2}+\frac{c+a+c+b}{2}\)

\(=\frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}\)

\(=\frac{\left(2a+a+a\right)+\left(2b+b+b\right)+\left(2c+c+c\right)}{2}\)

\(=\frac{4\cdot\left(a+b+c\right)}{2}=\frac{4\cdot2}{2}=4\)

Vậy \(maxP=4\Leftrightarrow a=b=c=\frac{2}{3}\)

 C=﴾x+1﴿^3+﴾x‐1﴿^3 ‐3x﴾x+1﴿﴾x‐1﴿

   =﴾x^ 3+3x^ 2+3x+1﴿+﴾x^ 3 ‐3x ^2+3x‐1﴿‐3x﴾x^ 2 ‐1﴿

   =x^ 3+3x^ 2+3x+1+x^ 3 ‐3x^ 2+3x‐1‐3x^ 3+3x

   =‐x^ 3+9x

a)\(P=4x^3-\left(2-4x\right).\left(x^2-3x+1\right)\)

\(=4x^3-\left(2x^2-6x+1-4x^2+12x^2-4x\right)\)

\(=4x^3-2x^2+6x-1+4x^2-12x^2+4x\)

\(=4x^3-10x^2+10x-1\)

b) Thay \(x=\frac{-1}{2}\) vào biểu thức trên

Ta Có : \(4.\left(\frac{-1}{2}\right)^3-10.\left(\frac{-1}{2}\right)^2+10.\left(\frac{-1}{2}\right)-1\)

       \(=\frac{-1}{2}-\frac{5}{2}-5-1\)

       \(=-3-5-1\)

\(=-8-1=-9\)

thanks  bạn ạ

Ko có gì

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

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

\(=\frac{x-1}{\sqrt{y}-1}.\frac{|y-2\sqrt{y}+1|}{|(x^2-2x+1)|}\)

\(=\frac{x-1}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}y\ge0\\x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}y\ge0\\x\ne1\end{cases}}}\)

\(A=\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x^2-2x+1\right)^2}}\)

\(=\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}}\)

\(=\frac{\left(x-1\right)|\sqrt{y}-1|}{\left(\sqrt{y}-1\right)\left(x-1\right)}=\frac{|\sqrt{y}-1|}{\left(\sqrt{y}-1\right)}\)

TH1 : \(y>1\Rightarrow\sqrt{y}>1\Rightarrow\sqrt{y}-1>0\)

\(\Rightarrow|\sqrt{y}-1|=\sqrt{y}-1\)

\(\Rightarrow A=\frac{\sqrt{y}-1}{\sqrt{y}-1}=1\)

Th2 : \(0< y< 1\Rightarrow\sqrt{y}< 1\Rightarrow\sqrt{y}-1< 0\)

\(\Rightarrow|\sqrt{y}-1|=-\left(\sqrt{y}-1\right)\)

\(\Rightarrow A=\frac{-\left(\sqrt{y}-1\right)}{\sqrt{y}-1}=-1\)

KL : Nếu \(0< y< 1\Rightarrow A=-1\)

Nếu \(y>1\Rightarrow A=1\)