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a) 5x2 -20
= 5(x2 -4)
=5 (x2 -22)
= 5(x-2)(x+2)
b) 16 - (x+y)2
=42 -(x+y)2
= (4-x-y)(4+x+y)
a, \(5\left(x^2-4\right)=5\left(x-2\right)\left(x+2\right)\)
b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)
mấy bài này áp dụng hđt là được nhé
\(x^2+x-20=\left(x^2+5x\right)-\left(4x+20\right)=x\left(x+5\right)-4\left(x+5\right)=\left(x+5\right)\left(x-4\right)\)
x2 + x - 20
= x2 + 5x - 4x - 20
= ( x2 + 5x ) - ( 4x + 20 )
= x( x + 5 ) - 4( x + 5 )
= ( x - 4 )( x + 5 )
\(\left(x^2-3x+2\right)\left(x^2-9x+20\right)-40=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-40\)
\(=\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40\)
Đặt \(t=x^2-6x+5\) thì ta có \(t\left(t+3\right)-40=t^2+3t-40=\left(t+8\right)\left(t-5\right)\)
Suy ra \(\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40=\left(x^2-6x+13\right)\left(x^2-6x\right)=x\left(x-6\right)\left(x^2-6x+13\right)\)
\(\Rightarrow\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\cdot\left(x^2+2x\right)+20=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
x20 + x + 1 = (x20 - x2) + (x2 + x + 1)
= x2(x18 - 1) + (x2 + x + 1)
= x2(x9 - 1)(x9 + 1) + (x2 + x + 1)
=(x11 + x)(x3 - 1)(x6 + x3 + 1) + (x2 + x + 1)
= (x17 + x14 + x11 + x7 + x4 + x)(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x18 + x15 + x12 + x8 + x5 + x2 - x17 - x14 - x11 - x7 - x4 - x + 1)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
\(a,x^2+9x+20=x^2+4x+5x+20.\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(c,x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2-2\right)-\left(2x\right)^2=\left(x^2-2x-2\right)\left(x^2+2x-2\right)\)
\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
\(x^2+x-20\)
\(=x^2+5x-4x-20\)
\(=x\left(x+5\right)-4\left(x+5\right)\)
\(=\left(x-4\right)\left(x+5\right)\)
cảm ơn