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\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)
Xem lại đề ý b
a) \(4x^3y^2-8x^2y+12xy^2=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x-2y\right)^2-49=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-6x+9\right)-25=\left(x-3\right)^2-25=\left(x-3-5\right)\left(x-3+5\right)=\left(x-8\right)\left(x+2\right)\)
a) 4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)
b) 3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)
c) x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)
d) x2−6x−16=(x2−6x+9)−25=(x−3)2−25=(x−3−5)(x−3+5)=(x−8)(x+2)
a) \(4x^3y^2-8x^2y+12xy^2=4xy.x^2y-4xy.2x+4xy.3y=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=\left(3x^2-6xy\right)-\left(5x-10y\right)=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x^2-4xy+4y^2\right)-49=\left(x-2y\right)^2-7^2=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-8x\right)+\left(2x-16\right)=x\left(x-8\right)+2\left(x-8\right)=\left(x-8\right)\left(x+2\right)\)
a: =5x(x-y)-7(x-y)
=(x-y)(5x-7)
b: =x(x+2y)+(x+2y)
=(x+2y)(x+1)
c; =(x-3)^2-9y^2
=(x-3-3y)(x-3+3y)
a
\(5x^2-5xy+7y-7x\\ =5x\left(x-y\right)+7\left(y-x\right)\\ =5x\left(x-y\right)-7\left(x-y\right)\\ =\left(5x-7\right)\left(x-y\right)\)
b
\(x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)
c
\(x^2-6x-9y^2+9\\ =x^2-6x+9-\left(3y\right)^2\\ =x^2-2.x.3+3^2-\left(3y\right)^2\\ =\left(x-3\right)^2-\left(3y\right)^2\\ =\left(x-3-3y\right)\left(x-3+3y\right)\)
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
Bài 1:
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)
1, =3x (2x -3y)
c, = 3x(x-y) -2(x-y)
= (3x-2)(x-y)
2, Ta có: x2 -6x+10= (x-3)2 +11
Nhận xét: (x-3)2 >= 0 với mọi số thực x
=> (x-3)2 +1 >= 1 >0 (đpcm)
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
