=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=...=2^32-1

nhân hết ra là xong:))

bài về nhà hs phải tự làm

Cái bước (2²-1)(2² + 1)(2⁴ +1)(2¹⁶+1) làm như thế nào mà ra vậy

        \(M=1.\left(a+b\right)\left(a^2+b^2\right).......\)

    \(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)....\)

   \(=\left(a^2-b^2\right)\left(a^2+b^2\right)....\)

                        \(=\left(a^4-b^4\right)\left(a^4+b^4\right)......\)

                                            \(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)

                                                             \(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\)

                                                              \(=a^{32}-b^{32}\)

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

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

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

\(=\left(2^8-1\right).\left(2^8+1\right)\left(2^{16}+1\right)....\left(2^{64}+1\right)+1\)

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

\(=2^{64}-1+1=2^{64}\)

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

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

Bài này có rắc rối đâu em?

Thực hiện phép tính trong ngoặc lại là ra dạng (n+1)/n.

1 dãy các số liên tục kéo dài nhân với nhau thì triệt tiêu là xong!

Chúc em học tốt!

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\)

\(\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)

\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)

\(\Rightarrow A=2-\dfrac{1}{2^{2017}}=\dfrac{2^{2018}-1}{2^{2017}}\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\)

\(2A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)\)

\(2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)

\(A=2-2^{2017}\)

đề hình như sai thì phải