a. Thay x = 4 vào biểu thức A , ta có:
\(A=\dfrac{4^2+12}{4^2-4}=\dfrac{16+12}{16-4}=\dfrac{28}{12}=\dfrac{7}{3}\)
b.
\(A-B=\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}-\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(A-B=\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}-\dfrac{4x+8}{\left(x-2\right)\left(x+2\right)}\)
\(A-B=\dfrac{x^2+12-4x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)\(=S\)
Đúng 2
Bình luận (0)
