Rút gọn biểu thức:P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
CM đẳng thức: (a+b+c)^3= a^3+b^3+c^3+3(a+b)(b+c)(c+a)
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6 tháng 8 2016
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
IM
6 tháng 8 2016
a)
Ta có
a chia 5 dư 4
=> a=5k+4 ( k là số tự nhiên )
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)
Vì 25k^2 chia hết cho 5
40k chia hết cho 5
16 chia 5 dư 1
=> đpcm
2) Ta có
\(12=\frac{5^2-1}{2}\)
Thay vào biểu thức ta có
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)
\(\Rightarrow P=\frac{5^{16}-1}{2}\)
3)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
AD
Rút gọn biểu thức
A= 12(52 +1)(54 +1)(58 +1)(516 +1)
Chứng minh
(a+b+c)3 = a3+b3+c3+3(a+b)(b+c)(c+a)
2
MD
31 tháng 5 2017
Bài 1:
A = \(12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> \(\left(5^2-1\right)A\) = \(12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> 24A = \(12\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{12}{24}.\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{32}-1\right)\)
MD
31 tháng 5 2017
Bài 2:
Ta có: \(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
= \(\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\) => đpcm
1 tháng 2 2022
a: \(P=\left(5x-1-5x-4\right)^2=\left(-3\right)^2=9\)
b: \(Q=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3\)
c: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{5^{32}-1}{2}\)
20 tháng 12 2021
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
5 tháng 9 2018
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
10 tháng 9 2018
Ta có :
\(x^3+y^3+z^3-3xyz\)
\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(\Rightarrow\left(x+y+z\right)\left[\left(x+y^2\right)-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
P/s tham khảo nha
hok tốt
(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
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