Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(=\left(x-2\right)^2\)
b) \(=\left(2x+1\right)^2\)
c) \(=\left(4x-3y\right)\left(4x+3y\right)\)
d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)
e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)
f) \(=\left(x-y\right)\left(x^2+xy+y^2\right)\)
g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)
h) \(=\left(x+2\right)^3\)
i) \(=\left(1-x\right)^3\)
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(4x^2+4x+1=\left(2x+1\right)^2\)
g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
`1)x^3-7x+6`
`=x^3-x-6x+6`
`=x(x-1)(x+1)-6(x-1)`
`=(x-1)(x^2+x-6)`
`=(x-1)(x^2-2x+3x-6)`
`=(x-1)[x(x-2)+3(x-2)]`
`=(x-1)(x-2)(x+3)`
`2)x^3-9x^2+6x+16`
`=x^3-2x^2-7x^2+14x-8x+16`
`=x^2(x-2)-7x(x-2)-8(x-2)`
`=(x-2)(x^2-7x-8)`
`=(x-2)(x^2-8x+x-8)`
`=(x-2)[x(x-8)+x-8]`
`=(x-2)(x-8)(x+1)`
`3)x^3-6x^2-x+30`
`=x^3+2x^2-8x^2-16x+15x+30`
`=x^2(x+2)-8x(x+2)+15(x+2)`
`=(x+2)(x^2-8x+15)`
`=(x+2)(x^2-3x-5x+15)`
`=(x+2)[x(x-3)-5(x-3)]`
`=(x+2)(x-3)(x-5)`
`4)2x^3-x^2+5x+3`
`=2x^3+x^2-2x^2-x+6x+3`
`=x^2(2x+1)-x(2x+1)+3(2x+1)`
`=(2x+1)(x^2-x+3)`
`5)27x^3-27x^2+18x-4`
`=27x^3-9x^2-18x^2+6x+12x-4`
`=9x^2(3x-1)-6x(3x-1)+4(3x-1)`
`=(3x-1)(9x^2-6x+4)`
1) Ta có: \(x^3-7x+6\)
\(=x^3-x-6x+6\)
\(=x\left(x^2-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
2) Ta có: \(x^3-9x^2+6x+16\)
\(=x^3-2x^2-7x^2+14x-8x+16\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x-8\right)\left(x+1\right)\)
3) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
Chia nhỏ ra cậu ơi :v
Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
a) x⁴ + 2x² + 1
= (x²)² + 2.x².1 + 1²
= (x² + 1)²
b) 4x² - 12xy + 9y²
= (2x)² - 2.2x.3y + (3y)²
= (2x - 3y)²
c) -x² - 2xy - y²
= -(x² + 2xy + y²)
= -(x + y)²
d) (x + y)² - 2(x + y) + 1
= (x + y)² - 2.(x + y).1 + 1²
= (x - y + 1)²
e) x³ - 3x² + 3x - 1
= x³ - 3.x².1 + 3.x.1² - 1³
= (x - 1)³
g) x³ + 6x² + 12x + 8
= x³ + 3.x².2 + 3.x.2² + 2³
= (x + 2)³
h) x³ + 1 - x² - x
= (x³ + 1) - (x² + x)
= (x + 1)(x² - x + 1) - x(x + 1)
= (x + 1)(x² - x + 1 - x)
= (x + 1)(x² - 2x + 1)
= (x + 1)(x - 1)²
k) (x + y)³ - x³ - y³
= (x + y)³ - (x³ + y³)
= (x + y)³ - (x + y)(x² - xy + y²)
= (x + y)[(x + y)² - x² + xy - y²]
= (x + y)(x² + 2xy + y² - x² + xy - y²)
= (x + y).3xy
= 3xy(x + y)
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
\(a,=y\left(y-2\right)\\ b,=3x\left(x^2-2x+1\right)=3x\left(x-1\right)^2\\ c,=\left(y-1\right)\left(27x^2+9x^3\right)=9x^2\left(x+3\right)\left(y-1\right)\\ d,=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\\ e,=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\\ f,=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\\ g,=\left(2-x\right)\left(x+1\right)\\ h,=\left(x-1\right)\left(3x-6\right)=3\left(x-1\right)\left(x-2\right)\)
a: =y(y-2)
b: \(=3x^2\left(x^2-2x+1\right)=3x^2\left(x-1\right)^2\)
d: \(=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\)
a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
a) x3+ 6x2+12x+8
=(x+2)3
b)x3-3x2+3x-1
=(x-1)3
c)1-9x+27x2-27x3
=(1-3x)3
d)x3 +\(\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
=(x+\(\frac{1}{2}\))3 ( phần này mik là khác đầu bài bạn đi 1 chút nhưng mik tôn trọng ý kiến của bạn hơn nên mik nghĩ mik làm sai)
e) 27x3-54x2y+36xy2-8y3
=(3x-2y)2
a) x3 + 6x2 + 12x + 8
= (x^3+2^3)+6x.(x+2)
= (x+2).(x^2-2x+4)+6x(x+2)
= (x+2).(x^2+4x+4)
b) x3 - 3x2 + 3x - 1
= (x^3-1) -3x.(x-1)
= (x-1).(x^2+x+1) - 3x(x-1)
= (x-1).(x^2-2x+1)
Câu d ko hiểu đề :v
e) 27x3- 54 x2y + 36 xy2 - 8y3
= (27x^3-8y^3)-(54x^2y+36xy^2)
= (3x-2y).(9x^2+6xy+4y^2)-18xy(3x-2y)
= (3x-2y).(9x^2-12xy+4y^2)
Thế nhé :)