ta có x^2 -4 = (x-2)(x+2)

đkxđ của C là x khác 2 và trừ 2

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

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

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

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

để C = 0 => x-1 = 0

=> x= 1 ( thỏa mãn điều kiện xác định)

c, để C dương 

=> x-1 dương 

=> x-1 >0

=> x>1

a) Để biểu thức xác định \(\Rightarrow\hept{\begin{cases}x^2-4\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)

\(\Rightarrow x\ne2;-2\)

Vậy ...

b) \(C=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)

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

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

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

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

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

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

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

Để C = 0 \(\Rightarrow x-1=0\) 

\(\Rightarrow x=1\)

Vậy ...

c) Để C > 0 thì \(x-1>0\Rightarrow x>1\)

Vậy ...

1. a, => -12x+60+21-7x = 5

=> 81 - 19x = 5

=> 19x = 81 - 5 = 76

=> x = 76 : 19 = 4

Tk mk nha

a) Để \(\frac{11}{\sqrt{x}-5}\)nhận giá trị nguyên thì \(\sqrt{\text{x}}-5\inƯ\left(11\right)\)(DK : \(0\le x\ne25\))

Vì \(\sqrt{\text{x}}-5\ge-5\)nên ta có : 

\(\sqrt{x}-5\in\left\{-1;1;11\right\}\)\(\Rightarrow\sqrt{x}\in\left\{4;6;16\right\}\Rightarrow x\in\left\{16;36;256\right\}\)

b) \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)(DK : \(0\le x\ne9\))

Để B nhận giá trị nguyên thì \(\sqrt{x}-3\inƯ\left(4\right)\)

Vì \(\sqrt{\text{x}}-3\ge-3\)nên ta có : 

\(\sqrt{\text{x}}-3\in\left\{-2;-1;1;2;4\right\}\)\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)

\(A=\left| x-2009\right|+\left|2010-x\right|>=\left|x-2009+2010-x\right|=1\)

Dấu = xảy ra khi 2009<=x<=2010

a)IxI>=0 =>2+IxI>=2 =>D_min=2 khi x= 0