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\(4A=4\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5^2-1\right)\left(5^2+1\right).....\left(5^{2048}+1\right)\left(hdt\left(a-b\right)\left(a+b\right)=a^2-b^2\right)=\left(5^4-1\right)\left(5^4+1\right)......\left(5^{2048}+1\right)=\left(5^8-1\right).....\left(5^{2048}+1\right)=.....=\left(5^{1024}+1\right)\left(5^{1024}-1\right)\left(5^{2048}+1\right)=\left(5^{2048}-1\right)\left(5^{2048}+1\right)=5^{4096}-1\)
Sửa đề
B = 2(3+1)(32+1)(34+1)(38+1)(316+1)
= (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
= (32-1)(32+1)(34+1)(38+1)(316+1)
= (34-1)(34+1)(38+1)(316+1)
= (38-1)(38+1)(316+1)
= (316-1)(316+1)
= (332-1)
Mình sửa đề bài nha:
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
\(=\frac{5^{32}-1}{2}\)
Chúc bạn học tốt!
\(A=3\left(2^3+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right).9\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{32}-1\right)\)
12
= \(\frac{24}{2}\)
= \(\frac{1}{2}\left(25-1\right)\)
= \(\frac{1}{2}\left(5^2-1\right)\)
Chép đề sai kìa
\(\left(2x+1\right)^2-\left(3x+2\right)^2\)
\(=\left(2x+1+3x+2\right).\left(2x+1-3x-2\right)\)
\(=\left(5x+3\right).\left(-x-1\right)\)
\(\left(x-y+1\right)\left(x+y-1\right)=\left[x-\left(y-1\right)\right]\left[x+\left(y-1\right)\right]=x^2-\left(y-1\right)^2\)
Trả lời :
\(\left(x-y+1\right).\left(x+y-1\right)\)
\(=\left[x-\left(y-1\right)\right].\left[x+\left(y-1\right)\right]\)
\(=x^2-\left(y-1\right)^2\)
Study well
I am a loser: Bạn chép đề sai nha, mình sửa luôn.
\(A=3\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^4-1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^8-1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^{16}-1\right)\cdot\left(2^{16}+1\right)\)
\(A=2^{32}-1\)
Vậy...