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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

 

 

6 tháng 8 2016

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)

 

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)

\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

 

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 = \(\dfrac{1}{2}\left(5^{32}-1\right)\)

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

17 tháng 8 2016

a)P= 15+45x+33x^2-55x^3

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: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{5^{32}-1}{2}\)

5 tháng 9 2018

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

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

7 tháng 12 2020

bạn viết thế này khó nhìn quá

26 tháng 11 2021

nhìn hơi đau mắt nhá bạn hoa mắt quá

17 tháng 10 2022

Bài 4: 

=>x(x^2+1)=0

=>x=0

Bài 5: 

=>\(3n^3+n^2+9n^2-1-4⋮3n+1\)

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

Bài 3: 

\(a^3+b^3+c^3-3bac\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)