\(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow2A=3^{128}-1\)

\(\Leftrightarrow A=\frac{3^{128}-1}{2}\)

ko hieu

(2x-7)^2-6(2x-7)(x-3)=0

=> (2x-7)(2x-7-6x+18)=0

=>(2x-7)(11-4x)=0

=>2x-7=0 hoặc 11-4x=0

+) Nếu 2x-7=0 thì 2x=7 nên x=7/2

+) Nếu 11-4x=0 thì 4x=11 nên x=11/4

 Nhớ k cho mình với nha, mình trả lời đầu tiên.

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)

\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0

Vậy x = -2004

Bài làm:

Ta có: \(x+y+z=8\Leftrightarrow\left(x+y+z\right)^2=64\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=64\)

Mà \(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\z^2+x^2\ge2zx\end{cases}}\)\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)

Thay vào ta có: \(64\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow xy+yz+zx\le\frac{64}{3}\)

Dấu "=" xảy ra khi: \(x=y=z=\frac{8}{3}\)

Vậy Max(B) = 64/3 khi x = y = z = 8/3