Tính
S=\(-1^2+2^2-3^2+4^2-...+2016^2\)
A=\(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}.4^{64}\)
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Những câu hỏi liên quan
Tính
S=\(-1^2+2^2-3^2+4^2-...+2016^2\)
A=\(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}.4^{64}\)
Mình đang gấp giúp mình nha!
Tìm x biết
\(a,-\frac{1}{2}\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
\(b,\sqrt{9\left(5x-1\right)}-\sqrt{16\cdot\left(5x-1\right)}+\sqrt{36\left(5x-1\right)}=15\)
MÌNH ĐANG CẦN GẤP GIẢI CỤ THỂ GIÚP MÌNH NHA
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
Giúp mình với mình cần gấp !!! Cảm ơn !
BT7: Tính
\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
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3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(C=5^{32}-1\)
4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)
\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)
\(D=4^{128}-1\)
5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)
\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)
....
\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)
\(E=5^{512}-1\)
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Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
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a/\(\left(1+2^4+2^8\right):\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{11}\right)\)
b/\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{2004}-1\right)\)
GIÚP MÌNH NHA CHIỀU MÌNH NỘP RỒI
Câu b: Đặt \(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{2004}-1\right)\)
Ta có: \(\frac{1}{2}-1=\left(-\frac{1}{2}\right);\frac{1}{3}-1=\left(-\frac{2}{3}\right);...;\frac{1}{2004}-1=\left(-\frac{2003}{2004}\right)\)
\(\Rightarrow B=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{2003}{2004}\right)\)
Vì B là 2003 thừa số âm nhân lại với nhau nên B là số âm
\(\Rightarrow B=-\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}\right)=-\frac{1}{2004}\)
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Câu a: Đặt \(A=1+2^4+2^8;B=1+2+2^2+...+2^{11}\)
\(\Rightarrow16A=2^4+2^8+2^{12}\) \(\Rightarrow15A=2^{12}-1\) \(\Rightarrow A=\frac{2^{12}-1}{15}\) \(\left(1\right)\)
\(\Rightarrow2B=2+2^2+2^3+...+2^{12}\) \(\Rightarrow B=2^{12}-1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A:B=\frac{2^{12}-1}{15}:\left(2^{12}-1\right)=\frac{1}{15}\)
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1) frac{2}{5}xfrac{1}{3}-frac{2}{15}:frac{1}{5}+frac{3}{5}xfrac{1}{3}2)left(3frac{1}{3}+2,5right):left(3frac{1}{6}-4frac{1}{5}right)-frac{11}{31}3)left[6+left(frac{1}{2}right)^3-left|-frac{1}{2}right|right]:frac{3}{12}4)frac{18}{37}+frac{8}{24}+frac{19}{37}-1frac{23}{24}+frac{2}{3}5)left(-2right)^3xleft(frac{3}{4}-0,25right):left(2frac{1}{4}-1frac{1}{6}right)6)left(frac{2}{5}right)^2+5frac{1}{2}xleft(45-2right)+frac{23}{4}7)frac{4}{9}-19frac{1}{3}-frac{4}{9}x39frac{1}{3}8)left(-frac{1}{2}right)^...
Đọc tiếp
1) \(\frac{2}{5}x\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}x\frac{1}{3}\)
2)\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)
3)\(\left[6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right]:\frac{3}{12}\)
4)\(\frac{18}{37}+\frac{8}{24}+\frac{19}{37}-1\frac{23}{24}+\frac{2}{3}\)
5)\(\left(-2\right)^3x\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
6)\(\left(\frac{2}{5}\right)^2+5\frac{1}{2}x\left(45-2\right)+\frac{23}{4}\)
7)\(\frac{4}{9}-19\frac{1}{3}-\frac{4}{9}x39\frac{1}{3}\)
8)\(\left(-\frac{1}{2}\right)^2:\frac{1}{4}-2x\left(-\frac{1}{2}\right)^2\)
9)\(125\%x\left(-\frac{1}{2}\right)^3:\left(1\frac{5}{16}-1,5\right)+2008^0\)
giúp mình nha mai mình học rùi
\(A=\left[1,5+2\frac{1}{2}-\left(\sqrt{8}\right)^2\right]:\left[4\frac{1}{2}-\sqrt{0,25}\right]-2\frac{3}{4}\)
Mình đang cần gấp
a)4.left(frac{1}{4}right)^2+25.left[left(frac{3}{4}right)^3:left(frac{5}{4}right)^3right]:left(frac{3}{2}right)^3b)2^3+3.left(frac{1}{2}right)^0+-1+left[left(-2right)^2:frac{1}{2}right]-8c) A1000-left{left(-5right)^3.left(-2right)^3-11.left[7^2-5.2^3+8left(11^2-121right)right]right}giúp mình với mọi người ơiai làm nhanh mà đúng đầu tiien mình tặng 3 tích
Đọc tiếp
a)\(4.\left(\frac{1}{4}\right)^2+25.\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3\)
b)\(2^3+3.\left(\frac{1}{2}\right)^0+-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8\)
c) A=\(1000-\left\{\left(-5\right)^3.\left(-2\right)^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
giúp mình với mọi người ơi
ai làm nhanh mà đúng đầu tiien mình tặng 3 tích