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Ta có: \(\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(=x^{32}-1\)
Bạn tham khảo nhé!
a)(x+1)(x+2)(x+3)(x+4)+1
=(x+1)(x+4)(x+2)(x+3)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4) thì (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
Thay: =(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16) thì (x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
Thay: =(x2+10x+16+4)2
=(x2+5x+20)2
a)(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4)
Ta có: (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
=(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16)
Ta có:(x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
=(x2+10x+16+4)2
=(x2+5x+20)2
Mk yêu bé Shin-Conan lém
quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6
=1\x-1\x+6
=6\x(x+6)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
\(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2-2x-3\right)\)
\(=2x-1\)
\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
đấy là tích không phải tổng nhé
Đặt \(A=3\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{32}-1\right)\)
\(\Leftrightarrow A=\frac{3\left(x^{32}-1\right)}{x^2-1}\)