Ta có: \(\dfrac{\left[\left(2x-2\right):2+1\right].2x+2}{2}=210\)

\(\Rightarrow\left[2\left(x-1\right):2+1\right].2\left(x+1\right)=420\)

\(\Rightarrow x-1+1.\left[2\left(x+1\right)\right]=420\)

\(\Rightarrow x.2\left(x+1\right)=420\)

\(\Rightarrow x\left(x+1\right)=210\)

\(\Rightarrow x\left(x+1\right)=14.15\)

\(\Rightarrow x=14\)

Vậy \(x=14\)

\(\dfrac{x-2}{4}=\dfrac{5+x}{3}\Leftrightarrow\left(x-2\right).3=\left(5+x\right).4\)

\(\Leftrightarrow3x-6=4x+20\)

\(\Leftrightarrow3x=4x+20-6\)

\(\Leftrightarrow3x=4x+14\)

\(\Leftrightarrow3x-4x=14\)

\(\Leftrightarrow\left(-1\right).x=14\)

\(\Leftrightarrow x=-14\)

mk nhầm cái đoạn \(\left(...5^{1997}\right)\) chuyển thành (...7) nha

\(A=\left(..3\right)^{1999}-\left(...5^{1997}\right)=\left(...3^4\right)^{499}.3^3-\left(...7^4\right)^{499}.7\)

\(A=\left(...1\right)^{499}.\left(...7\right)-\left(...1\right)^{499}.7\)

\(A=\left(...1\right).7-\left(...1\right).7=\left(...7\right)-\left(...7\right)=\left(...0\right)\)

\(\Rightarrow A⋮5\)

lấy bài bd