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tính biểu thức:
P=1+1/2(1+2)+1/3+(1+2+3)+.....+1/16(1+2+3+...+16)
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PT
Phạm Tiến Quang
26 tháng 11 2021
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Những câu hỏi liên quan
17 tháng 6 2016
(a+b+c)3=(a+b)3+3(a+b)2c+3(a+b)c2+c3
=a3+b3+3ab.(a+b)+3(a+b)2c+3(a+b)c2+c3
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[a.(b+c)+c.(b+c)]
=a3+b3+c3+3(a+b)(b+c)(c+a)
=>dpcm
17 tháng 6 2016
P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(52-1)(52+1)(54+1)(58+1)(516+1)
=(54-1)(54+1)(58+1)(516+1)
=(58-1)(58+1)(516+1)
=(516-1)(516+1)
=532-1
==>P=(532-1)/2
29 tháng 7 2021
Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
29 tháng 7 2021
Rút gọn: (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3-1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(32-1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(34-1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(38-1)(38 + 1)(316 + 1)(332 + 1)
A=(316-1)(316 + 1)(332 + 1)
A=(332 - 1)(332 + 1)
A=364-1
=>A=(364-1) /2
MH
14 tháng 9 2021
c)\(\left(xy^2-1\right)\left(x^2y+5\right)\)
\(=x^3y^3+5xy^2-x^2y-5\)
d)\(4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x^2+1\right)\)
\(=4\left(x^2-\dfrac{1}{4}\right)\left(4x^2+1\right)\)
\(=4\left(4x^4+x^2-x-\dfrac{1}{4}\right)\)
\(=16x^4+4x^2-4x-1\)
12 tháng 7 2016
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\).Ta có :
\(=>\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
...............................................................................
Cuối cùng \(=>2A=3^{64}-1\).
\(=>A=\frac{3^{64}-1}{2}\)
12 tháng 7 2016
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=...........................................\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
11 tháng 7 2021
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)