Gọi số cần tìm là a, ta có:

\(a\times20-a\times15=25\\ a\times\left(20-15\right)=25\\ a\times5=25\\ a=5\)

Kết quả đúng của phép tính đó là:

\(5\times20=100\)

\(42+x:3=51\\ x:3=9\\ x=3\)

\(315-300:\left(x:15-7\right)=300\\ 300:\left(x:15-7\right)=15\\ x:15-7=20\\ x:15=27\\ x=405\)

\(2x\left(x+81\right)=0\\ \Rightarrow\left\{{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

Dạ e quên.

Nếu bài có nhầm gì thì mn góp ý nhé!

\(\dfrac{1}{3.10}+\dfrac{1}{10.17}+\dfrac{1}{17.24}+...+\dfrac{1}{73.80}-\dfrac{1}{2.9}-\dfrac{1}{9.16}-\dfrac{1}{16.23}-\dfrac{1}{23.30}\\ =\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{24}+...+\dfrac{1}{73}-\dfrac{1}{80}-\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+\dfrac{1}{23.30}\right)\\ =\dfrac{1}{3}-\dfrac{1}{80}-\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{30}\right)\\ =\dfrac{77}{240}-\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\\ =\dfrac{77}{240}-\dfrac{7}{15}\\ =\dfrac{-7}{48}\)

C