Question

Bày em câu 2 vs ạ

Answer 1

ui tròi xuống thử chỗ lớp 6 dạo môn toán đi , khó hơn 8 lun:<

Answer 2

Câu 2 

1/ a, ĐKXĐ : \(x^2-1\ne0;1-x^2\ne0;\left(x^2+1\right)\left(x-1\right)\ne0\)

x khác 1 , -1 , 

Answer 3

Câu 2:

a) Điều kiện xác định: \(x\ne\pm1\)

\(A=\dfrac{x^2}{x^2-1}-\dfrac{2x+1}{1-x^2}-\dfrac{x^2+1}{\left(x^2+1\right)\left(x-1\right)}\)

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

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

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

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

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

\(=\dfrac{x+2}{x-1}\)

b) Ta có: \(x^2+3x+2=0\)

\(\Leftrightarrow x^2+x+2x+2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

Với \(x=-1\) thay vào \(A=\dfrac{x+2}{x+1}\) ta có:

\(A=\dfrac{-1+2}{-1+1}=0\)

Với \(x=-2\) thay vào \(A=\dfrac{x+2}{x+1}\) ta có:

\(A=\dfrac{-2+2}{-2+1}=0\)