\(A=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\)

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\) ( do \(3=2^2-1\) nhé bạn)

\(A=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\) ( do \((2^2-1)(2^2+1)=2^4-1\) Đây là hằng đẳng thức \(a^2-b^2=(a-b)(a+b)\) 

\(A=\left(2^8-1\right)\left(2^8+1\right)...\left(2^{256}+1\right)+1\)

....

\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)

\(A=2^{512}-1+1=2^{512}\)

Vậy \(A=2^{512}\)

\(x^2-x+y^2+y+\frac{1}{2}=0\)

\(\left(x^2-x+\frac{1}{4}\right)+\left(y^2+y+\frac{1}{4}\right)=0\)

\(\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=0\)

Ta có: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\forall x\\\left(y+\frac{1}{2}\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\ge0\forall}x;y\)

Mà \(\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=0\)

\(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+\frac{1}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)

hello mik biết giải bài này nhưng bn phải viết rõ

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)

\(=\)\(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\)\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x+11=t\) ta có : 

\(=\)\(\left(t-1\right)\left(t+1\right)-8\)

\(=\)\(t^2-1-8\)

\(=\)\(t^2-9\)

\(=\)\(\left(t-3\right)\left(t+3\right)\)

\(=\)\(\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)\)

\(=\)\(\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

Chúc bạn học tốt ~ 

\(a)\)\(3x^2-6xy+3y^2-12\)

\(=\)\(3\left(x^2-2xy+y^2\right)-12\)

\(=\)\(3\left(x-y\right)^2-12\)

\(=\)\(3\left[\left(x-y\right)^2-4\right]\)

\(=\)\(3\left(x-y-4\right)\left(x-y+4\right)\)

\(b)\)\(x^2+5x+6\)

\(=\)\(\left(x^2+2x\right)+\left(3x+6\right)\)

\(=\)\(x\left(x+2\right)+3\left(x+2\right)\)

\(=\)\(\left(x+2\right)\left(x+3\right)\)

Chúc bạn học tốt ~ 

a) 3x²-6xy+3y²-12=3(x²-2xy+y²)-12=3(x-y)²-12=3[(x-y)²-4]=3(x-y-2)(x-y+2)

b)x²+3x+2x+6=x(x+3)+2(x+3)=(x+3)(x+2)