\(-x^2+3x+5=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=5\\x_1x_2=\dfrac{c}{a}=-5\end{matrix}\right.\)
Ta có : \(A=\dfrac{x_1+2}{x_2-2}+\dfrac{x_2+2}{x_1-2}\)
\(=\dfrac{\left(x_1+2\right)\left(x_1-2\right)+\left(x_2+2\right)\left(x_2-2\right)}{\left(x_2-2\right)\left(x_1-2\right)}\)
\(=\dfrac{\left(x_1^2-4\right)\left(x_2^2-4\right)}{x_1x_2-2\left(x_1+x_2\right)+4}\)
\(=\dfrac{\left(x_1x_2\right)^4-4\left(x_1^2+x_2^2\right)+16}{x_1x_2-2\left(x_1+x_2\right)+4}\)
\(\)\(=\dfrac{\left(x_1x_2\right)^4-4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+16}{x_1x_2-2\left(x_1+x_2\right)+4}\)
\(=\dfrac{\left(-5\right)^4-4\left(5^2-2\left(-5\right)\right)+16}{-5-2.5+4}\)
\(=-\dfrac{501}{11}\)
giúp mik với ạ chỉ cần rút gọn 6 cái biểu thức thui ạ .mong mn giúp đỡ.
