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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
x^3+x^2-2x-8
= (x-2)(x^2+3x+4)
nah bạn chúc bạn học tốt nha
x3 + x2 - 2x - 8
= ( x3 - 8 ) + ( x2 - 2x )
= ( x - 2 ) . ( x2 + 2x + 4 ) + x ( x - 2 )
= ( x - 2 ) .( x2 + 2x + 4 + x )
= ( x-2 ) . ( x2 + 3x + 4 )
=(x^3+6x^2+12x+8)+y^3
=(x^3+3x^2+3x2^2+2^3)+y^3
=(x+2)^3+y^3
=(x+2+y)((x+2)^2-(x+2)y+y^2)
=(x+2+y)(x^2+4x+4-xy-2y+y^2)
=(x+2+y)(x^2+y^2-xy+4x-2y+4)
x3 - 6x2 + 12x - 8
= x3 - 2x2 - 4x2 + 4x + 8x - 8
= (x3 - 2x2) - (4x2 - 8x) + (4x - 8)
= x2.(x - 2) + 4x.(x - 2) + 4.(x - 2)
= (x - 2).(x2 + 4x + 4)
= (x - 2).(x2 + 2x + 2x + 4)
= (x - 2).[x.(x + 2) + 2.(x + 2)]
= (x - 2).(x + 2).(x + 2)
= (x - 2).(x + 2)2
x^9 + x^8 + x^7 - x^3 + 1
= x^7 ( x^2 + x + 1 ) - ( x^3 - 1 )
= x^7 ( x^2 + x + 1 ) - ( x - 1 )(x^2 + x + 1 )
= ( x^7 - x + 1 )(x^2 + x + 1 )
(x+1)(x+2)(x+3)(x+4)-8
=[(x+1).(x+4)].[(x+2).(x+3)]-8
=(x2+5x+4).(x2+5x+6)-8
Đặt (x2+5x+4)=t =>(x2+5x+6)=t+2
Thay vào biểu thức ta có:
(x2+5x+4).(x2+5x+6)-8
t.(t+2)-8
=t2+2t+1-9
=(t+1)2-32
=(x2+5x+4+1)-32
=(x2+5x+5+3).(x2+5x+5-3)
=(x2+5x+8).(x2+5x+2)
=
ta làm như sau :
\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8.\)
\(\Rightarrow\left(x^2+5X+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+4=t\)
\(\Leftrightarrow t\left(t+2\right)-8\)
\(\Leftrightarrow t^2+2t-8\Leftrightarrow t^2+2t+1-9\)
\(\Leftrightarrow\left(t+1\right)^2-3^2\)
\(\Leftrightarrow\left(t-2\right)\left(t+4\right)\)
\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+4\right)\)