Question

[akakak21]

Help mik với cần gấp ạ

Answer 1

[Nguyễn Huy Tú]

Bài 1 : 

a, ĐK : \(x\ne\pm2\)

b,  \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+3}{4-x^2}\)

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

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

c, Vì x = -2 ktm đkxđ (loại)

Thay x = 3 ta được : \(\frac{9+1}{1.5}=\frac{10}{5}=2\)

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

\(\Rightarrow3x^2+3=-\left(x^2-4\right)=4-x^2\Leftrightarrow4x^2-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)( tmđk )

e, Để A ko âm khi \(\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}>0\Rightarrow\left(x-2\right)\left(x+2\right)>0\)

TH1 : \(\hept{\begin{cases}x-2>0\\x+2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x>-2\end{cases}}\Leftrightarrow x>2}\)

TH2 : \(\hept{\begin{cases}x-2< 0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2\\x< -2\end{cases}\Leftrightarrow x< -2}\)

Vậy x > 2 ; x < -2 thì A ko âm