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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: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
a) \(x^2-9+2\left(x+3\right)=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x-3+2\right)=\left(x+3\right)\left(x-1\right)\)
b) \(x^2-10x+25-3\left(x-5\right)=\left(x-5\right)^2-3\left(x-5\right)=\left(x-5\right)\left(x-5-3\right)=\left(x-5\right)\left(x-8\right)\)
c) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)
\(a,=x\left(x+y\right)+5\left(x+y\right)=\left(x+5\right)\left(x+y\right)\\ b,=x\left(y-x\right)-3\left(y-x\right)=\left(x-3\right)\left(y-x\right)\\ c,=18x-4x^3=2x\left(9-2x^2\right)\\ d,=\left(x-2\right)^2-4y^2=\left(x-2y-2\right)\left(x+2y-2\right)\\ e,=x^2-x-9x+9=\left(x-1\right)\left(x-9\right)\\ f,=4x^2-6x+2x-3=\left(2x-3\right)\left(2x+1\right)\)
a.
$64x^3-16x^2+x=x(64x^2-16x+1)$
$=x(8x-1)^2$
b.
$36-4xy+24y-x^2=(4y^2+24y+36)-(x^2+4xy+4y^2)$
$=(2y+6)^2-(x+2y)^2=(2y+6-x-2y)(2y+6+x+2y)$
$=(6-x)(x+4y+6)$
c.
$x^2+10x-2010.2020$
$=x^2+10x-(2015-5)(2015+5)
$=x^2+10x-(2015^2-5^2)$
$=(x^2+10x+5^2)-2015^2=(x+5)^2-2015^2$
$=(x+5-2015)(x+5+2015)=(x-2010)(x+2020)$
d.
$25x^2-121+22y-y^2$
$=(5x)^2-(y^2-22y+11^2)$
$=(5x)^2-(y-11)^2=(5x-y+11)(5x+y-11)$
e.
$(x^2+2x)(x^2+2x-2)-3$
$=(x^2+2x)^2-2(x^2+2x)-3$
$=(x^2+2x)^2+(x^2+2x)-3(x^2+2x)-3$
$=(x^2+2x)(x^2+2x+1)-3(x^2+2x+1)$
$=(x^2+2x+1)(x^2+2x-3)$
$=(x+1)^2[x(x-1)+3(x-1)]$
$=(x+1)(x-1)(x+3)$
\(a,=\left(5x-1\right)^2\\ b,=\left(x+4\right)^2\\ c,=\left(4x+3y\right)^2\\ d,=\left(\dfrac{x}{4}+2y\right)^2\)
a, \(x^2-5x+6=x^2+x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)
b, \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[x\left(x-2\right)+5\left(x-2\right)\right]=3\left(x-2\right)\left(x+5\right)\)
c, \(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)
a, x2 - 10x + 24 = x2 - 2.x.5 + 25 - 1 = (x-5)2 - 1 = (x - 5 - 1).(x - 5 + 1) = (x - 6).(x - 4)
b, x2 - x - 30 = x2 - 6x + 5x - 30 = x.(x - 6) + 5.(x - 6) = (x - 6).(x + 5)
c, x2 + 9x + 8 = x2 + x + 8x + 8 = x.(x + 1) + 8.(x + 1) = (x + 1).(x + 8)
Câu d đề có phải sai không vậy bạn, mình giải theo đề này nha: x2 - 10x + 16 = x2 - 8x - 2x +16 = x.(x - 8) - x.(x - 8) = (x - 2).(x - 8)