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a ) ( x2 + 2x + 5 )( x2 + 2x + 3 ) - 8
= ( x2 + 2x + 5 )[ ( x2 + 2x + 5 ) - 2 ] - 8
= ( x2 + 2x + 5 )2 - 2 . ( x2 + 2x + 5 ) + 1 - 9
= ( x2 + 2x + 5 - 1 )2 - 9
= ( x2 + 2x + 4 )2 - 33
= ( x2 + 2x + 4 - 3 )( x2 + 2x + 4 + 3 )
= ( x2 + 2x + 1 )( x2 + 2x + 7 )
b ) ( x2 + 2x )( x2 + 2x - 2 ) - 3
= ( x2 + 2x )[ ( x2 + 2x ) - 2 ] - 3
= ( x2 + 2x )2 - 2 . ( x2 + 2x ) + 1 - 4
= ( x2 + 2x - 1 )2 - 22
= ( x2 + 2x - 1 - 2 )( x2 + 2x - 1 + 2 )
= ( x2 + 2x - 3 )( x2 + 2x + 1 )
= ( x2 + 2x - 3 )( x + 1 )2
trả lời :
- \(\left(x^2+2x+5\right)\left(x^2+2x+3\right)\)
Đặt: \(x^2+2x+5=t\Rightarrow x^2+2x+3=t+2\),ta có:
\(t\left(t+2\right)-8\)
\(=t^2+2t-8\)
\(=t^2+4t-2t-8\)
\(=t\left(t+4\right)-2\left(t+4\right)\)
\(=\left(t+4\right)\left(t-2\right)\)
Thay vào cách đặt , ta có:
\(\left(x^2+2x+5+4\right)\left(x^2+2x+5-2\right)\)
\(=\left(x^2+2x+9\right)\left(x^2+2x+3\right)\)
\(=\left(x^2+2x+9\right)\left(x^2+3x-x+3\right)\)
\(=\left(x^2+2x+9\right)\left(x+3\right)\left(x-1\right)\)
- \(\left(x^2+2x\right)\left(x^2+2x-2\right)-3\)
Đặt : \(x^2+2x=t\Rightarrow\left(x^2+2x-2\right)=t-2\),ta có:
\(t\left(t-2\right)-3\)
\(=t^2-2t-3\)
\(=t^2-3t+t-3\)
\(=t\left(t-3\right)+\left(t-3\right)\)
\(=\left(t-3\right)\left(t+1\right)\)
Thay vào cách đặt, ta có:
\(\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+3x-x-3\right)\left(x+1\right)^2\)
\(=\left(x+3\right)\left(x-1\right)\left(x+1^2\right)\)
#hok tốt #
Đây là cách hiện đại :
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-\left(2x^3-2x\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x^2+1\right)-2x\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)-2x\right)\)
a,=\(x^4-x^3-x^3+x^2-x^2+x+x-1\)
cu hai so nhom 1 nhom roi dat thua so chung la xong
b,x^4+x^3+x^3+x^2+x^2+x+x+1
cu hai so lai nhom 1 nhom va dat thua so chung
a,\(x^2y-4y=y\left(x^2-4\right)=y\left(x-2\right)\left(x+2\right)\)
b,\(x^2-y^2-2x+1=\left(x^2-2x+1\right)-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x-y-1\right)\)
c,\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
x2y - 4y = y( x2 - 4 ) = y( x - 2 )( x + 2 )
x2 - y2 - 2x + 1 = ( x2 - 2x + 1 ) - y2 = ( x - 1 )2 - y2 = ( x - 1 - y )( x - 1 + y )
5x2 + 5xy - x - y = ( 5x2 + 5xy ) - ( x + y ) = 5x( x + y ) - ( x + y ) = ( x + y )( 5x - 1 )
1. = (x^2-x+6+x-3).(x^2-x+6-x+3) [ áp dụng a^2-b^2=(a-b).(a+b)]
= (x^2+3).(x^2-2x+9)
2. Vì 105 lẻ => 2x+5y+1 và 2^|x| + x^2+x+y lẻ
Mà 2y chẵn , 1 lẻ => 5y chẵn => y chẵn
Lại có : x^2+x=x.(x+1) chẵn
=> 2^|x| lẻ => x=0
Khi đó : (5y+1).(y+1) = 105
Đến đó bạn tự tìm ước của 105 rùi giải đi
k mk nha
x^4 - 2x^3 - 2x^2 - 2x - 3
=x^4 + x^3 - 3x^3 - 3x^2 + x^2 + x - 3x - 3
=x^3(x+1) - 3x^2(x+1) + x(x+1 ) - 3(x+1)
=(x+1)(x^3 - 3x^2 + x - 3)
=(x+1)[x^2 (x - 3) + x - 3]
=(x+1)(x - 3)(x^2 + 1)
x^2 - y^2 - 2x + 4y - 3
= (x^2 - 2x + 1) - (y^2 - 4y + 4)
= (x + 1)^2 - (y + 2)^2
= (x + 1 - y - 2)(x + 1 + y + 2)
= (x - y - 1)(x + y + 3)