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\(4A=4\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5^2-1\right)\left(5^2+1\right).....\left(5^{2048}+1\right)\left(hdt\left(a-b\right)\left(a+b\right)=a^2-b^2\right)=\left(5^4-1\right)\left(5^4+1\right)......\left(5^{2048}+1\right)=\left(5^8-1\right).....\left(5^{2048}+1\right)=.....=\left(5^{1024}+1\right)\left(5^{1024}-1\right)\left(5^{2048}+1\right)=\left(5^{2048}-1\right)\left(5^{2048}+1\right)=5^{4096}-1\)
x4 + x3 + 2x2 + 1
= (x4 + 2x2 + 1) + x3
= (x2 + 1)2 + x3
còn bài nào ko??
56457675675758768364576567568768963454256364576756
\(x^4+x^3+2x^2+1\)
\(=\left(x^4+2x^2+1\right)+x^3\)
\(=\left(x^2+1\right)^2+x^3\)
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...
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)
\(P=\frac{2x-1}{x^2-2}\left(ĐKXĐ:x\ne\pm\sqrt{2}\right)\)
\(\Leftrightarrow Px^2-2P=2x-1\)
\(\Leftrightarrow Px^2-2x-2P+1=0\)
*Nếu P = 0 thì ....
*Nếu P khác 0 thì pt trên là bậc 2
\(\Delta'=1-P\left(2P+1\right)=-2P^2-P+1\)
Có nghiệm thì \(\Delta'\ge0\Leftrightarrow-1\le P\le\frac{1}{2}\)
Nên Pmin = -1
Đến đây dạng này khi biết kết quả thì phân tích dễ r ha , từ làm nốt câu còn lại nhé , tương tự luôn
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!
\(A=3\left(2^3+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right).9\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{32}-1\right)\)