a, ĐKXĐ: x khác -2 và 2
b, nếu \(x\ge1\) thì x=1(TMĐK)\
thay vào A =-1
nếu x<1 thì x=-2 (KTMĐK)
Sửa đề tí:
\(A=\frac{3}{x-2}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
ĐKXD: x khác 2 và -2
\(A=\frac{3}{x-2}-\frac{3}{x+2}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x+6}{\left(x-2\right)\left(x+2\right)}-\frac{3x-6}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{12+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(|x|+1=1\Leftrightarrow x=0\)
\(A=\frac{12+0}{\left(-2\right)\left(2\right)}=\frac{12}{-4}=-3\)
ĐKXĐ: x khác +-2
\(A=\frac{-3}{x-2}-\frac{3}{x+2}+\frac{3x^2}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{-3.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}-\frac{3.\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{3x^2}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right).\left(x+2\right)}=\frac{-6x+3x^2}{\left(x-2\right).\left(x+2\right)}=\frac{3x.\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}=\frac{2x}{x+2}\)
rút gọn rồi, thay vào |x+1|=1 => x= mấy rồi tính
\(a,A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\)
\(=\frac{-3}{x-2}+\frac{-3}{x+2}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(-3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{-3x-6-3x+6+3x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{-6x+3x^2}{\left(x-2\right)\left(x+2\right)}=\frac{3x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3x}{x+2}\)
\(\RightarrowĐKXĐ:x\ne-2\)
( mk chưa học kĩ lắm, cs sai sót mong bn thông cảm )
=) shit bo đề ko sai đâu =))
\(2-x=-x+2=-\left(x-2\right)\)
bn đưa dấu âm lên trên tử thức (số 3)
a) ĐKXĐ của A là \(x\ne\pm2\)
\(A=\frac{3}{2-x}-\frac{3}{x+2}+\frac{3x^2}{x^2-4}\\ =\frac{-3\left(x+2\right)}{\left(x-2\right)}+\frac{-3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3x^2}{\left(x+2\right)\left(x-2\right)}\\ =\frac{-3x-6-3x+6+3x^2}{\left(x+2\right)\left(x-2\right)}=\frac{-6x+3x^2}{\left(x+2\right)\left(x-2\right)}\\ =\frac{3x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x}{x+2}\)
\(|x+1|=1\Rightarrow\hept{\begin{cases}x+1=1\\x+1=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-2\end{cases}}\)
Vì x=0 thỏa mãn đk x khác 2;-2 nên thay x=0 vào A
\(\Rightarrow A=\frac{3.0}{0+2}=0\)
Vậy...................