Phân tích các đa thức sau thành nhân tử
a)3x5y3-12x4y4+12x3y5-48x3y3
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\(a,=5\left(a-4b\right)\\ b,=\left(y+1\right)^2-x^2=\left(y+1-x\right)\left(x+y+1\right)\)
a) \(9x^2-16\)
\(=\left(3x\right)^2-4^2\)
\(=\left(3x-4\right)\left(3x+4\right)\)
b) \(x^2+4xy+4y^2-3x-6y\)
\(=\left(x^2+4xy+4y^2\right)-\left(3x+6y\right)\)
\(=\left[x^2+2\cdot x\cdot2y+\left(2y\right)^2\right]-3\left(x+2y\right)\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)
#\(Toru\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`3x^2 + 6xy + 3y^2 - 3z`
`= 3*x^2 + 3*2xy + 3y^2 - 3z`
`= 3(x^2 + 2xy + y^2 - z)`
`b,`
`x^3 + x^2y - x^2z - xyz`
`= x(x + y)(x-z)`
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
\(a,x-xy+y-y^2\\=(x-xy)+(y-y^2)\\=x(1-y)+y(1-y)\\=(1-y)(x+y)\\---\\b,x^2-4x-y+4(?)\\---\\c,x^2-2x-3\\=x^2+x-3x-3\\=x(x+1)-3(x+1)\\=(x+1)(x-3)\)
Bạn xem lại đề câu b nhé!
\(x^4-27x=x\left(x^3-27\right)=x\left(x-3\right)\left(x^2+3x+9\right)\)
\(27x^5+x^2=x^2\left(27x^3+1\right)=x^2\left[\left(3x\right)^3+1^3\right]=x^2\left(3x+1\right)\left(9x^2-3x+1\right)\)
a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)
b) \(6x^2+x-2=2x\left(3x+2\right)-1\left(3x+2\right)=\left(3x+2\right)\left(2x-1\right)\)
\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)
a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)
b: \(ax^2+a-a^2x-x\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)
a) 3x5y3-12x4y4+12x3y5-48x3y3
=(3x5y3-12x4y4)+(12x3y5-48x3y3)
=3x3y3(3x5y3-12x4y4)+(12x3y5-48x3y3)
=(x2-4xy)+(4y2-16)
(Nếu có sai mong bạn thông cảm nhá!!!)