kết quả là -2

a. Để P được xđ thì MT phải khác 0.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9\ne0\\x^2+3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ne0\\x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.\)

b. \(P=\left(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\right).\dfrac{x-3}{x+3}\)

\(P=\left(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\right).\dfrac{x-3}{x+3}\)

\(P=\left(\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right).\dfrac{x-3}{x+3}\)

\(P=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)

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

\(P=\dfrac{1}{x}\)

\(n_{NH_4NO_3}=\dfrac{m}{M}=\dfrac{20}{80}=0,25\left(mol\right)\)

\(\Rightarrow n_N=2.n_{NH_4NO_3}=2.0,25=0,5\left(mol\right)\)

\(m_N=n.M=0,5.14=7\left(kg\right)\)

8520=8000+500+2

tick mik nha bạn

muốn tính diện tích hình thang=(a+b)xh:2 (đáy bé+đáy lớn)x chiều cao :2

tick mình nha!!!!

hình tam giác rồi đề như thế nào ấy

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

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

b) \(\dfrac{8-x}{\left(x+2\right)\left(x-3\right)}+\dfrac{2}{x+2}\)

\(=\dfrac{8-x}{\left(x+2\right)\left(x-3\right)}+\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(=\dfrac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}\)

\(=\dfrac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}=\dfrac{x+2}{\left(x+2\right)\left(x-3\right)}\)

\(=\dfrac{1}{x-3}\)