\(a,A=1+3+3^2+...+3^{125}\\ \Rightarrow3A=3+3^2+3^3+...+3^{126}\\ \Rightarrow2A=3^{126}-1\\ \Rightarrow A=\dfrac{3^{126}-1}{2}\\ c,2A=3^{2x}-1\\ \Rightarrow3^{126}-1=3^x-1\\ \Rightarrow x=126\)

\(d,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{124}+3^{125}\right)\\ A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{124}\left(1+3\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{124}\right)\\ A=4\left(1+3^2+...+3^{124}\right)⋮4\)

a) \(2^{4-2x}=2^{16}\)(biến đổi \(16^4\Rightarrow2^{16}\))

=> 4 - 2x = 16

          2x = 4 - 16

          2x = -12

          x   = -12 : 2

          x   = -6

a) 16^4=2^16

=> 4-2x=16=> x=-6

Xem lại câu b nha bạn!

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.

Vui lòng viết yêu cầu bài :>

a, (x-2)(3x+5)=(2x-4)(x+1)
<=> (x-2)(3x+5)-2(x-2)(x+1)=0
<=>(x-2)(3x+5-2x-2)=0
<=>(x-2)(x+3)=0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

b, (2x+5)(x-4)=(x-5)(4-x)
<=>(2x+5)(x-4)-(x-5)(4-x)=0
<=>(x-4)(2x+5+x-5)=0
<=>(x-4)3x=0
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}}\)