Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Phải là (2+1)(2²+1)(2⁴+1)...(2³²+1)- 2^64
(2+1)(2²+1)(2⁴+1)...(2³²+1)
=(2-1)(2+1)(2²+1)(2⁴+1)...(2³²+1)
=(2²-1)(2²+1)(2⁴+1)...(2³²+1)
=(2⁴-1)(2⁴+1)...(2³²+1)=…=2^64-1
Vậy C=-1
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
rút gọn biểu thức
a)2x(2x−1)2−3x(x+3)(x−3)−4x(x+1)2
=2x(4x2-4x+1)-3x.(x2-9)-4x(x2+2x+1)
=8x3-8x2+2x-3x3-27x-4x3-8x2-4x
=8x3-16x2-7x3-29x
Bài 11:
1) Sửa lại đề là: \(A=127^2+146.127+73^2\)
\(\Rightarrow A=127^2+2.127.73+73^2\)
\(\Rightarrow A=\left(127+73\right)^2\)
\(\Rightarrow A=200^2\)
\(\Rightarrow A=40000\)
Vậy \(A=40000.\)
2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)
\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(\Rightarrow B=18^8-\left(18^8-1\right)\)
\(\Rightarrow B=18^8-18^8+1\)
\(\Rightarrow B=0+1\)
\(\Rightarrow B=1\)
Vậy \(B=1.\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow D=\frac{3^{32}-1}{2}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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.
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{32}+1\right)-2^{64}\)
\(=2^{64}-1-2^{64}=-1\)
a)
$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$
$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$
$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$
$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$
$=2005^2-1002.2005=2005(2005-1002)=2011015$
b)
$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$
$=(2^{32}-1)(2^{32}+1)-2^{64}$
$=2^{64}-1-2^{64}=-1$
c) Do $x=16$ nên $x-16=0$
$R(x)=x^4-17x^3+17x^2-17x+20$
$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$
$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$
$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$
d) Do $x=12$ nên $x-12=0$. Khi đó:
$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$
$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$
$=(x-12)(x^9-x^8+x^7-....+x)-x+10$
$=0-x+10=-x+10=-12+10=-2$