\(\left(x+5\right)^4=\left(x+5\right)^6\)
\(\Rightarrow\left(x+5\right)^4-\left(x+5\right)^6=0\)
\(\Rightarrow\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Rightarrow\)\(x=-4\)
Câu 2 : không có yêu cầu đề bài thì làm kiểu gì?
\(\left(x+5\right)^4=\left(x+5\right)^6\)
\(\Rightarrow\left(x+5\right)^6-\left(x+5\right)^4=0\)
\(\Rightarrow\left(x+5\right)^4.\left[\left(x+5\right)^2-1\right]=0\)
\(\Rightarrow\left(x+5\right)^4=0\)hoặc \(\left(x+5\right)^2-1=0\)
=> x + 5 = 0 hoặc (x + 5)2 = 1
=> x + 5 = 0 hoặc x + 5 = 1 hoặc x + 5 = -1
=> x = -5 hoặc x = -4 hoặc x = -6
\(\left(x+5\right)^4=\left(x+5\right)^6\)
\(\left(x+5\right)^6-\left(x+5\right)^4=0\)
\(\left(x+5\right)^4.\left[\left(x+5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+5\right)^4=0\\\left(x+5\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-4\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=-5\\x=-4\end{cases}}\)
Sửa đề chút:
a)\(A=3+3^2+3^3+...+3^{100}\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-3\)
\(A=\frac{3^{101}-3}{2}\)
b) Ta có: \(2A+3=3^n\)
\(\Rightarrow2.\left(\frac{3^{101}-3}{2}\right)+3=3^n\)
\(3^{101}-3+3=3^n\)
\(3^{101}=3^n\)
\(\Rightarrow n=101\)
Vậy \(n=101\)
Tham khảo nhé~
