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\(M=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)+1\)
\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
\(=\left(a^2+5a+4\right)^2+2\left(a^2+5a+4\right)+1\)
\(=\left(a^2+5a+5\right)^2\) là bình phương của 1 số nguyên (đpcm)
M=(x+1)(x+4)(x+2)(x+3)+1
=(x2+5x+4)(x2+5x+6)+1
dat x2+5x+5=a ta co
M=(a+1)(a-1)+1
=a2-1+1
=a2
thay a boi x2+5x+5 ta co M=(x2+5x+5)2 (1)
ma x la so nguyen nen x2+5x+5 la so nguyen (2)
tu (1) va (2) thi M la binh phuong cua 1 so nguyen
1. \(M=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(=\left[\left(a+1\right)\left(a+4\right)\right]\left[\left(a+2\right)\left(a+3\right)\right]+1\)
\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
\(=\left(a^2+5a+4\right)^2+2\left(a^2+5a+4\right)+1\)
\(=\left(a^2+5a+5\right)^2\)
=> Đpcm
M = ( a + 1 )( a + 2 )( a + 3 )( a + 4 ) + 1
= [ ( a + 1 )( a + 4 ) ][ ( a + 2 )( a + 3 ) ] + 1
= [ a2 + 5a + 4 ][ a2 + 5a + 6 ] + 1
Đặt t = a2 + 5a + 4
M <=> t[ t + 2 ] + 1
= t2 + 2t + 1
= ( t + 1 )2
= ( a2 + 5a + 4 + 1 )2 = ( a2 + 5a + 5 )2 ( đpcm )
( x2 + x + 1 )( x2 + x + 2 ) - 12 (*)
Đặt t = x2 + x + 1
(*) <=> t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + x + 1 - 3 )( x2 + x + 1 + 4 )
= ( x2 + x - 2 )( x2 + x + 5 )
= ( x2 + 2x - x - 2 )( x2 + x + 5 )
= [ x( x + 2 ) - 1( x + 2 ) ]( x2 + x + 5 )
= ( x + 2 )( x - 1 )( x2 + x + 5 )
tranvantoancv.violet.vn/present/showprint/entry_id/11064865
A=x^4+6x^3+7x^3-6x+1=x^4+6(x^3-2x^2)+(9x^2-6x+1)=x^4+2x^2(3x-1)+(3x-1)^2=(x^2+3x-1)^2
2A = (3+1)(3-1)(3^2+1)(3^4+1)...(3^64+1)
2A= (3^2-1)(3^2+1)(3^4+1)...(3^64+1)
Cứ tiếp tục như thế ta dc
2A= 3^128 -1
A = (3^128-1)/2
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=8.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
.....
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(=3^{128}-1\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
bài 2 :
x3+7y=y3+7x
x3-y3-7x+7x=0
(x-y)(x2+xy+y2)-7(x-y)=0
(x-y)(x2+xy+y2-7)=0
\(\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\left(loại\right)\\x^{2^{ }}+xy+y^2-7=0\end{matrix}\right.\)
x2+xy+y2=7 (*)
Giải pt (*) ta đc hai nghiệm phan biệt:\(\left[{}\begin{matrix}x=1va,y=2\\x=2va,y=1\end{matrix}\right.\)
M = ( a + 1 )( a + 2 )( a + 3 )( a + 4 ) + 1
= [ ( a + 1 )( a + 4 ) ][ ( a + 2 )( a + 3 ) ] + 1
= ( a2 + 5a + 4 )( a2 + 5a + 6 ) + 1
Đặt t = a2 + 5a + 4
M = t( t + 2 ) + 1
= t2 + 2t + 1
= ( t + 1 )2
= ( a2 + 5a + 4 + 1 )2
= ( a2 + 5a + 5 )2
Vì a nguyên => a2 + 5a + 5 nguyên
Vậy M = ( a + 1 )( a + 2 )( a + 3 )( a + 4 ) + 1 là bình phương của một số nguyên ( đpcm )