Phân tích đa thức sau thành nhân tử:
a) x^2-4 +(x-2)^2.
b) x^3-2x^2+x-xy^2.
c) x^3-4x^2-12x+27.
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Lời giải:
a. Không phân tích được nữa
b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
a) `x^4+2x^3-4x-4`
`=(x^4-4)+(2x^3-4x)`
`=(x^2-2)(x^2+2)+2x(x^2-2)`
`=(x^2-2)(x^2+2+2x)`
b) `x^3-4x^2+12x-27`
`=(x^3-27)-(4x^2-12x)`
`=(x-3)(x^2+3x+9)-4x(x-3)`
`=(x-3)(x^2+3x+9-4x)`
`=(x-3)(x^2-x+9)`
c) `xy-4y-5x+20`
`=y(x-4)-5(x-4)`
`=(y-5)(x-4)`
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+2x^3-4x\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c) Ta có: \(xy-4y-5x+20\)
\(=y\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(y-5\right)\)
câu a đặt chung x ra là xong
câu b
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé
a) x3 - 2x2 + x - xy2
= x (x2 - 2x + 1 - y2)
= x [(x2 - 2x + 1) - y2]
= x [(x - 1)2 - y2]
= x [(x - 1) + y] [(x - 1) - y]
= x (x - 1 + y) (x - 1 - y)
b) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x3 + 33) - 4x (x + 3)
= (x + 3) (x2 - 3x + 32) - 4x (x + 3)
= (x + 3) [(x2 - 3x + 9) - 4x]
= (x + 3) (x2 - 3x + 9 - 4x)
= (x + 3) (x2 - 7x + 9)
#Học tôt!!!
~NTTH~
a) x2 – 4 + (x – 2)2
= (x2 – 22) + (x – 2)2 = (x – 2)(x + 2) + (x – 2)2
= (x – 2) [(x + 2) + (x – 2)]
= (x – 2)(x + 2 + x – 2)
= 2x(x – 2)
b) x3 – 2x2 + x – xy2
= x(x2 – 2x + 1 – y2) = x[(x2 – 2x + 1) – y2]
= x[(x – 1)2 – y2]
= x[(x – 1) + y] [(x – 1) – y]
= x(x – 1 + y)(x – 1 – y)
c) x3 – 4x2 – 12x + 27
= (x3 + 27) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9 – 4x)
= (x + 3)(x2 – 7x + 9)
\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)
a) = (x3 +33) -4x(x+3)
= (x+3)(x2 -3x+9-4x)
= (x+3)(x2 - 7x +9)