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1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
d)
x3 + 2x2y+ xy2 - 9x
=x*(x2+2xy+y2 -9)
=x*[ (x+y)2 -32 ]
=x * (x+y-3) * (x+y-3)
a, x4 + 2x3 +x2 = x4 +x3 +x3 +x2 =(x4+x3 )+(x3 +x2 ) =x3(x +1 ) + x2 (x+1 ) =(x+1)(x3+x2)
a) x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)2
= [x(x + 1)]2
= (x2 + x)2
b) 5x3 - 10xy + 5y2 - 20z2
= 5(x3 - 2xy + y2 - 4z2)
c) x2y - xy2 + x3 - y3
= xy(x - y) + (x - y)(x2 + xy + y2)
= (x - y)(x2 + 2xy + y2)
= (x - y)(x + y)2
d) x2 - xy + 4x - 2y + 4
= (x2 + 4x + 4) - (xy + 2y)
= (x + 2)2 - y(x + 2)
= (x + 2)(x + 2 - y)
d) x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3) + 2(x - 3)
= (x + 2)(x - 3)
f) 3x2 - 5x - 8
= 3x2 + 3x - 8x - 8
= 3x(x + 1) - 8(x + 1)
= (3x - 8)(x + 1)
g) x3 + 3x2 + 6x + 4
= (x3 + 3x2 + 3x + 1) + (3x + 3)
= (x + 1)3 + 3(x + 1)
= (x + 1)[(x + 1)2 + 3]
h) 3x3 - 5x2 - 6x + 8
= 3x3 - 3x2 - 2x2 - 6x + 8
= 3x3 - 3x2 - 2x2 + 2x - 8x + 8
= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)
= (3x2 - 2x - 8)(x - 1)
a) \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(x^2y-xy^2+x^3-y^3\)
\(=xy\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\)
\(=\left(x+2\right)^2-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-y+2\right)\)
e) \(x^2-x-6=\left(x+2\right)\left(x-3\right)\)
f) \(3x^2-5x-8\)
\(=\left(3x^2+3x\right)-\left(8x+8\right)\)
\(=3x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-8\right)\)
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
a )
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
b )
\(-5x^2-16x-3\)
\(=-5x^2-15x-x-3\)
\(=-5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(-5x-1\right)\left(x+3\right)\)
c )
\(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)-5\right]\)
d )
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
P/s : Mình bổ sung :
a )
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
d )
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
Mấy thánh đăng sớm z
a.
\(\left(3x+1\right)^2-\left(x+1\right)^2\\ =2x\left(4x+2\right)\)
b.https://hoc24.vn/hoi-dap/question/531769.html
Mình làm bài này r ; mà nó dài nên k mún làm lại
c.
\(5x^2-10xy+5y^2-20z^2\\ =5\left(x^2-2xy+y^2-4z^2\right)\\ =5\left(\left(x-y\right)^2-4z^2\right)\\ =5\left(x-y-2z\right)\left(x-y+2z\right)\)
m)
\(x^4+4\\ =\left(x^4+4x^2+4\right)-4x^2\\ =\left(x^2+2\right)^2-4x^2\\ =\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)